Solutions are homogeneous mixtures of two or more than two components. By homogenous mixture we mean that its composition and properties are uniform throughout the mixture. Generally, the component that is present in the largest quantity is known as solvent. Solvent determines the physical state in which solution exists. One or more components present in the solution other than solvent are called solutes. In this Unit we shall consider only binary solutions.

Solubility
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. It depends upon the nature of solute and solvent as well as temperature and pressure.
like dissolves like.
Dissolution - When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution.
Crystallisation- Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation.
Dynamic equilibrium - A stage is reached when the two processes occur at the same rate. Under such conditions, number of solute particles going into solution will be equal to the solute particles separating out and a state of dynamic equilibrium is reached. Solute + Solvent ⇌ Solution
Saturated solution - A saturated solution is a solution that has dissolved as much solute as it is capable of dissolving. In a saturated solution, no more solute can be dissolved at a given temperature. Unsaturated solution- Unsaturated solutions are solutions in which the amount of dissolved solute is less than the saturation point of the solvent (at that specific temperature gradient). As seen above If the amount of dissolved solute is equal to the saturation point of the solvent, the solution is called a saturated solution. What is the difference between a saturated and unsaturated solution? -
A saturated solution contains the maximum amount of solute that can dissolve in a solvent at a specific temperature and pressure, while an unsaturated solution can dissolve more solute.
⇒ Saturated solutions can be found in many everyday products and substances. a. Sugar water: A common example of a saturated solution is sugar water, where the maximum amount of sugar that can be dissolved in water at a particular temperature has been dissolved.
b. Carbonated drinks: Carbonated drinks like soda are saturated solutions of carbon dioxide gas in water under high pressure. The solubility of carbon dioxide in water decreases as temperature increases, so the drink will lose its carbonation if left at room temperature.
c. Silver chloride: Silver chloride is a sparingly soluble salt that forms a saturated solution in water. It is used in photography and silver plating.
d. Potassium nitrate: Potassium nitrate is a salt that forms a saturated solution in water. It is used as a fertilizer, food preservative, and in the production of gunpowder.
⇔ Factors affecting solubility vary on the state of the solute:-
a. Liquids in Liquids
b. Solids in Liquids
c. Gases in Liquids
Liquids In Liquids
Water is known as a universal solvent as it dissolves almost every solute except for a few. Certain factors can influence the solubility of a substance.
Based on the concentration of solute dissolves in a solvent, solutes are highly soluble, sparingly soluble or insoluble.
If a concentration of 0.1 g or more of a solute can be dissolved in a 100ml solvent, it is said to be soluble. While a concentration below 0.1 g is dissolved in the solvent it is said to be sparingly soluble.
Thus, it is said that solubility is a quantitative expression and expressed by the unit gram/ litre (g/L).
Based on solubility, different types of solution can be obtained. A saturated solution is a solution where a given amount of solute is completely soluble in a solvent at given temperature. On the other hand, a supersaturated solution is those where solute starts to salting out or precipitate after a particular concentration is dissolved at the same temperature.
Factors Affecting Solubility: -
Solubility of a Solid in a Liquid-
The solubility of a substance depends on the physical and chemical properties of that substance. In addition to this, other factors are Temperature, pressure and the type of bonds formed. Effect of temperature The solubility of a solid in a liquid is significantly affected by temperature changes. Consider the equilibrium represented Solute + Solvent ⇌ Solution
This, being dynamic equilibrium, must follow Le Chateliers Principle. In general, if in a nearly saturated solution, the dissolution process is endothermic (Δsol H > 0), the solubility should increase with rise in temperature and if it is exothermic (Δsol H < 0) the solubility should decrease. These trends are also observed experimentally.
⇒ Le Chatelier’s principles, also known as the equilibrium law, are used to predict the effect of some changes on a system in chemical equilibrium (such as the change in temperature or pressure). The principle is named after the French chemist Henry Louis Le Chatelier.
Le Chatelier discovered that equilibrium adjusts the forward and backward reactions in such a way as to accept the changes affecting the equilibrium conditions.
When factors like concentration, pressure, temperature, and inert gases that affect equilibrium are changed, the equilibrium will shift in that direction, where the effects caused by these changes are nullified.
⇔ Effect of pressure
Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.
Solubility of a Gas in a Liquid
The solubility of a gas in a liquid depends on temperature, the partial pressure of the gas over the liquid, the nature of the liquid and the nature of the gas.
Factors affecting solubiliy of gas in liquid
a. Nature of the gas and the solvent b. Effect of temperature(Decreases with increase in temperature) c. Effect of pressure
Henry’s law-
The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.
If we use the mole fraction of a gas in the solution as a measure of its solubility, then it can be said that the mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.
The most commonly used form of Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as: P ∝ x (or)
P = KH.x Here KH is the Henry’s law constant.
Henry's constant KH = 1/K
where K is the constant of proportionality whose value depends on the nature of the gas, the nature of solvent and the temperature.
Characteristics of Henry's constant
It is the function of the nature of the gas Greater the value of KH , lower is the solubility of the gas at the same partial pressure at a particular temperature The value of KH increases with the increase of temperature implying that the solubility decreases with the increase of the temperature at the same pressure.
Examples of Henry’s Law
Carbonated Drinks Henry’s law comes into play every time a bottle of carbonated drink is opened. The gas above the unopened carbonated drink is usually pure carbon dioxide, kept at a pressure which is slightly above the standard atmospheric pressure. As a consequence of Henry’s law, the solubility of carbon dioxide in the unopened drink is also high. When the bottle is opened, the pressurized CO2 escapes into the atmosphere (which is usually accompanied by a hissing sound). As the partial pressure of CO2 in the atmosphere above the drink rapidly decreases, the solubility of the carbon dioxide in the drink also decreases (due to Henry’s law). This causes the dissolved CO2 to come to the surface of the drink in the form of tiny bubbles and escape into the atmosphere.
If the carbonated drink is left open long enough, the concentration of carbon dioxide in the drink will reach an equilibrium with the concentration of carbon dioxide in the atmosphere (~0.05%), causing it to go flat (the drink loses its ‘fizzy’ taste).
Applications of Henry's law
In the production of carbonated beverages
In the deep sea diving
In the function of lungs
For climbers or people living at higher altitudes
Limitations of Henry's law
a. The pressure should be low and the temperature should be high i.e the gas should behave like an ideal gas b. The gas should not undergo compound formation with the solvent or association Colligative properties Certain properties of dilute solutions containing non-volatile solute do not depend upon the nature of the solute dissolved but depend only upon the concentration i.e., the number of particles of the solute present in the solution. Such properties are called colligative properties. The four well known examples of the colligative properties are, (1) Lowering of vapour pressure of the solvent. (2) Osmotic pressure of the solution. (3) Elevation in boiling point of the solvent. (4) Depression in freezing point of the solvent. Lowering of vapour pressure The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid. The vapour pressure of a liquid depends on, (1) Nature of liquid : Liquids, which have weak intermolecular forces, are volatile and have greater vapour pressure. For example, dimethyl ether has greater vapour pressure than ethyl alcohol. (2) Temperature : Vapour pressure increases with increase in temperature. This is due to the reason that with increase in temperature more molecules of the liquid can go into vapour phase. (3) Purity of liquid : Pure liquid always has a vapour pressure greater than its solution. Raoult's law : When a non-volatile substance is dissolved in a liquid, the vapour pressure of the liquid (solvent) is lowered. According to Raoult's law (1887), at any given temperature the partial vapour pressure (PA)of any component of a solution is equal to its mole fraction (XA) multiplied by the vapour pressure of this component in the pure state (p⁰A). That is, PA= PA⁰ X XA The vapour pressure of the solution (Ptotal) is the sum of the partial pressures of the components, i.e., for the solution of two volatile liquids with vapour pressures PA and PB Ptotal = PA+PB = (PA⁰ X XA)+( PB⁰ X XB) Alternatively, Raoult's law may be stated as "the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fradion of the solute in the solution." Relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure of the pure solvent. It is determined by Ostwald-Walker method. Thus according to Raoult's law, • Raoult's law is applicable only to very dilute solutions. • Raoult's law is applicable to solutions containing nonvolatile solute only. • Raoult's law is not applicable to solutes which dissociate or associate in the particular solution. Relation between Dalton's law and Raoult's law: The composition of the vapour in a equilibrium with the solution can be calculated applying Dalton's law of partial pressures. Let the mole fractions of vapours A and B be YA and YB respectively and total pressure P Ideal solutions 1. Obey Raoult's law at every range of concentration. 2. ΔHmix = 0; neither heat is evolved nor absorbed during dissolution. 3. ΔVmix = 0; total volume of solution is equal to sum of volumes of the components. 4. 5. - A, A - B, B - B interactions should be same, i.e.,A and B are identical in shape, size and character; Escaping tendency of 'A' and 'B' should be same in pure liquids and in the solution. 6. Examples: Dilute solutions; benzene + toluene: n-hexane + n-heptane; chlorobenzene + bromobenzene; ethyl bromide + ethyl iodide; n-butyl chloride + n-butyl bromide Non-ideal• solutions Positive deviation from Raoult's law 1. Do not obey Raoult's law. 2. ΔHmix > 0 Endothermic dissolution; heat is absorbed. 3. ΔVmix > 0 Volume is increased after dissolution. 4. 5. A - B attractive force should be weaker than A - A and B - B attractive forces. 'A' and 'B' have different shape, size and character. 'A' and 'B' escape easily showing higher vapour pressure than the expected value. Examples: Acetone +ethanol; acetone + CS2 ; water + methanol; water + ethanol; CCl4 +toluene; CCl4 +CHCl3; acetone + benzene; cyclohexane + ethanol Non-ideal solutions Negative deviation from Raoult's law 1. Do not obey Raoult's law. 2. ΔHmix < 0. Exothermic dissolution; heat is evolved. 3. ΔVmix mix < 0. Volume is decreased during dissolution. 4. 5. A - B attractive forces should be greater than A – A and B-B attractive forces. ' A' and 'B' have different shape, size and character. Escaping tendency of both components 'A' and 'B' is lowered showing lower vapour pressure than expected ideally. Examples: Acetone + aniline; acetone + chloroform; CH3OH + CH3COOH; H2O+HNO3 chloroform + diethyl ether water + HCI; acetic acid + pyridine; chloroform + benzene Differentiate a. Ideal solutions b. Non-ideal• solutions -Positive deviation from Raoult's law c. Non-ideal solutions -Negative deviation from Raoult's law Acetone +ethanol; acetone + CS2; water + methanol; water + ethanol; CCl4 +toluene; CCl4 +CHCl3; acetone + chloroform; CH3OH + CH3COOH; water + HCI; acetone + benzene; cyclohexane + ethanol Acetone + aniline; H2O+HNO3 chloroform + diethyl ether acetic acid + pyridine; chloroform + benzene Azeotropic mixture Azeotropes are defined as the mixtures of liquids which boil at constant temperature like a pure liquid and possess same composition of components in liquid as well as in vapour phase. Azeotropes are also called constant boiling mixtures because whole of the azeotropes changes into vapour state at constant temperature and their components cannot be separated by fractional distillation. Azeotropes are of two types as described below, (1) Minimum boiling azeotrope : For the solutions with positive deviation there is an intermediate composition for which the vapour pressure of the solution is maximum and hence, boiling point is minimum. At this composition the solution distils at constant temperature without change in composition. This type of solutions are called minimum boiling azeotrope. e.g., H2O +C2H5OH, ; CHCl3 + C2H5OH , (CH3)2CO + CS2 (2) Maximum boiling azeotrope : For the solutions with negative deviations there is an intermediate composition for which the vapour pressure of the solution is minimum and hence, boiling point is maximum. At this composition the solution distils at constant temperature without the change in composition. This type of solutions are called maximum boiling azeotrope. e.g., H2O +HCI, H2O +HNO3, H2O +HClO4. Abnormal molecular masses Molecular masses can be calculated by measuring any of the colligative properties. The relation between colligative properties and molecular mass of the solute is based on following assumptions. (1) The solution is dilute, so that Raoult's law is obeyed. (2) The solute neither undergoes dissociation nor association in solution. In case of solutions where above assumptions are not valid we find discrepencies between observed and calculated values of colligative properties. These anomalies are primarily due to (i) Association of solute molecules. (ii) Dissociation of solute molecules. Association of solute molecules: Certain solutes in solution are found to associate. This eventually leads to a decrease in the number of molecular particles in the solutions. Thus, it results in a decrease in the values of colligative properties. Colligative property α 1/ molecular mass of solute therefore, higher values are obtained for molecular masses than normal values for unassociated molecules. (ii) Dissociation of solute molecules : A number of electrolytes dissociate in solution to give two or more particles (ions). Therefore, the number of solute particles, in solutions of such substances, is more than the expected value. Accordingly, such solutions exhibit higher values of colligative properties. Since colligative properties are inversely proportional to molecular masses, therefore, molecular masses of such substances as calculated from colligative properties will be less than their normal values. Effects of Association/Dissociation Association is the joining of two or more particles to form one entity. An example of the association of two particles is the dimerization of carboxylic acids when dissolved in benzene. Dissociation refers to the splitting of a molecule into multiple ionic entities. For example, sodium chloride (NaCl) dissociates into Na+ and Cl– ions when dissolved in water. The effects of the association or dissociation of a solute on the solution, its colligative properties, and the Van’t Hoff factor are given below. The abnormality in the molecular mass can be explained as follows: The dissociation of solute molecules into multiple ions results in an increase in the number of particles. This, in turn, increases the colligative properties of the solution. Since the molar mass is inversely proportional to the colligative properties, its value tends to be lower than expected. When solute particles associate with each other, the total number of particles in the solution decreases, leading to a decrease in the colligative properties. In this case, the molar mass values obtained are higher than expected. Van't Hoff's factor (i) : In 1886, Van't Hoff introduced a factor 'i' called Van't Hoff’s factor, to express the extent of association or dissociation of solutes in solution. It is the ratio of the normal and observed molecular masses of the solute, i.e., i = Normal molecular mass / Observed molecular mass In case of association, observed molecular mass being more than the normal, the factor i has a value less than 1. But in case of dissociation, the Van't Hoffs factor is more than 1 because the observed molecular mass has a lesser value than the normal molecular mass. In case there is no dissociation the value of 'I' becomes equal to one. Introduction of the Van't Hoff factor modifies the equations for the colligative properties as follows- Degree of dissociation (α) : It is defined as the fraction of total molecules which dissociate into simpler molecules or ions- Degree of association (a) : It is defined as the fraction of the total number of molecules which associate or combine together resulting in the formation of a bigger molecules. Under what condition van’t Hoff factor ‘i’ is equal to unity? When non-electrolyte ‘solute’ does not undergo association or dissociation. Q2 Under what condition van’t Hoff factor ‘i’ is less than one? When solute undergoes association. Q3 What is the van’t Hoff factor for a compound which undergoes tetramerisation in an organic solvent? On adding HCl and ZnCl2 to both compounds, 2-methyl-2-propanol gives violet colouration while 1-propanol does not. Q4 Arrange the following in increasing order of acidity: phenol, ethanol, and water. i = Moles of compound after association/Moles of compound before association i= 1/4 = 0.25 Q5 How much molecular mass of NaCl is obtained experimentally using colligative properties? NaCl → Na+ + Cl– i=2 Van’t Hoff Factor(i)= Calculated Molar Mass/Observed molar mass So, Observed or experimentally obtained molar mass= Calculated Molar Mass/Van’t Hoff Factor(i) =(23 +35.5)/2 = 29.25 Q6 What is the van’t Hoff factor for a compound which undergoes dimerisation in an organic solvent? i = Moles of compound after association/Moles of compound before association i= 1/2 = 0.5

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E1. Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of (CH₆O₂) by mass.
E2. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.
E3. Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.
E4. If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.48 kbar.
E5. Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and, (ii) mole fractions of each component in vapour phase.
E6. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol^-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?
E7. 18 g of glucose, C₆H₁₂O₆, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol^-1.
E8. The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg mol^-1.
E9. 45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.
E10. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.
E11. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3bar. Calculate the molar mass of the protein.
E12. 2 g of benzoic acid (C₆H₅COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^-1. What is the percentage association of acid if it forms dimer in solution?
E13. 0.6 mL of acetic acid (CH₃COOH), having density 1.06 g mL^-1 , is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.

N1. Calculate the mass percentage of benzene (C₆H₆) and carbontetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
(so1Ref)- N1
N2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
(so1Ref)- N2
N3. Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO₃)₂. 6H₂O in 4.3 L of solution (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.
(so1Ref)- N3

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N4. Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
(so1Ref)- N4

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N5. 1.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.
(so1Ref)- N5

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N6. H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
(so1Ref)- N6

N7. Henry’s law constant for CO₂ in water is 1.67×10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

(so1Ref)- N7

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N8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
(so1Ref)- N8

N9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
(so1Ref)- N9

N10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
(so1Ref)- N10

N11. Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol^-1.
(so1Ref)- N11

N12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
(so1Ref)- N12

B1. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Homogeneous mixtures of two or more than two components are known as solutions.
There are three types of solutions.
(i) Gaseous solution
(ii) Liquid solution
(iii) Solid solution Gaseous solution
The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.
(ii) Liquid solution:
The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid.
For example, a solution of ethanol in water is a liquid solution.
(iii) Solid solution:
The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.
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B2. Give an example of a solid solution in which the solute is a gas.
Solution of Palladium in Hydrogen is an example of a solid solution in which the solute is a gas. In this example, the solid solvent is Palladium and the gaseous solute is Hydrogen.

B3. Define the following terms:
(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
(031202.004) (031202.005)

B4. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?
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B5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL^-1, then what shall be the molarity of the solution?
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B6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃and NaHCO₃ containing equimolar amounts of both?
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B7. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

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B8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, then what shall be the molarity of the solution?
(031202.0)

B9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
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B10. What role does the molecular interaction play in a solution of alcohol and water?

In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. Between the molecules of alcohol as well as water, hydrogen bonding is one of the interactive forces between the molecules. When alcohol and water are mixed together, new H-bonds also develop between alcohol and water molecules but these are weaker than the previous ones. So, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult's law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.
B11. Why do gases always tend to be less soluble in liquids as the temperature is raised?

Gas + Liquid ⇌ Dissolving gas; ΔH=−ve. Dissolution of gas is an exothermic process. As the temperature is raised, the equilibrium shifts in reverse direction (Le-Chatelier's principle). It results in decrease of solubility of gases in liquid.

⇒ Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. ⇔

B12. State Henry’s law and mention some important applications.
Henry's law states that the partial pressure of a gas in the vapour phase is proportional to a mole fraction of the gas in the solution.
P=KH × C
where C= concentration or mole fraction of dissolved gas, KH= Henry's law constant and P= partial pressure of dissolved gas.
Limitation of Henry's law
It has been observed that Henry's law is valid if
(i) pressure is low. At high pressure, the law becomes less accurate and the proportionality constant shows considerable deviations.
(ii) the temperature is not too low.
(iii) the gas is not highly soluble and
(iv) the gas neither reacts chemically with the solvent nor dissociates or associates in the solvent.
Applications of Henry’s law-

a. To increase the solubility of CO₂, in soft drinks and soda water, the bottle is sealed under high pressure.
b. Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
c. At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

B13. The partial pressure of ethane over a solution containing 6.56 × 10^-3g of ethane is 1 bar. If the solution contains 5.00 × 10^-2g of ethane, then what shall be the partial pressure of the gas?

B14. What is meant by positive and negative deviations from Raoult's law and how is the sign of DmixH related to positive and negative deviations from Raoult's law?

Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆sol H is positive. Similarly ∆solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute. So there is expansion in volume on solution formation. Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆sol H is negative.

B15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

B16. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

B17. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

B18. Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

B19. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

B20. A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

B21. Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C6H6), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.

B22. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

B23. Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I₂ and CCl₄ (iii) NaClO₄ and water (iv) methanol and acetone (v) acetonitrile (CH₃CN) and acetone (C₃H₆O).

B24. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.

B25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.
(i) Phenol: Partially soluble (Reason: Phenol has polar -OH group and non-polar C6H5 - group) (ii) Toluene: Insoluble (Reason: Toluene is non-polar, water is polar). (iii) Formic acid (HCOOH) : Highly soluble (Reason: Hydrogen bonding) (iv) Ethylene glycol: Highly soluble (Reason: H - bonds are formed although polarity is present) (v) Chloroform: Insoluble (Reason: H - bonds are formed although polarity is present). Chloroform is a slightly polar compound. But it cannot make any strong bond (like hydrogen bond) with water. Due to the lack of strong interaction with water its solubility in water is very less (or not soluble). (vi) Pentanol: Partially soluble (Reason: -OH group is polar but long hydrocarbon part is non-polar).

B26. If the density of some lake water is 1.25g mL^-1and contains 92 g of Na⁺ ions per kg of water, calculate the molarity of Na⁺ ions in the lake.
B27. If the solubility product of CuS is 6 × 10^-16, calculate the maximum molarity of CuS in aqueous solution.

B28. Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O<sub>4 is dissolved in 450 g of CH3CN.

B29. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 ´ 10^-3m aqueous solution required for the above dose.

B30. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

B31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

B32. Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10^-3, Kf = 1.86 K kg mol^-1.

B33. 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

B34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

B35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

B36. 100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

B37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is: 100 x xacetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 pacetone /mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 pchloroform /mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

B38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

B39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.

B40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.

B41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated.

SVO: 01‐ 19
O1. What are solutions?
Solutions are homogeneous mixtures of two or more than two components.
O2. What is homogeneous mixture?
Homogenous mixture is the one in which composition and properties are uniform throughout the mixture.
O3. What are components? Also specify solvent and solute.
The substances making up the solution are called components of the solution. Generally, the component that is present in the largest quantity is known as solvent. Solvent determines the physical state in which solution exists. One or more components present in the solution other than solvent are called solutes.
04. Give various types of solutions.
Types of Solutions Binary - having two components Ternary - having three components Quaternary - having four components Type of Solution Solute Solvent Common Examples Gaseous Solutions Gas Gas Mixture of oxygen and nitrogen gases Liquid Gas Chloroform mixed with nitrogen gas Solid Gas Camphor in nitrogen gas Liquid Solutions Gas Liquid Oxygen dissolved in water Liquid Liquid Ethanol dissolved in water Solid Liquid Glucose dissolved in water Solid Solutions Gas Solid Solution of hydrogen in palladium Liquid Solid Amalgam of mercury with sodium Solid Solid Copper dissolved in gold
O5. What is concentration of a solution?
The concentration of a solution may be defined as the amount of solute present in the given quantity of the solution. It can be described both qualitatively and quantitatively. When we say qualitatively it means solution is dilute or concentrated.
O6. Make a list of 7 ways to describe the concentration of a solution.
1. mass percentage or volume percentage or mass by volume percentage 2. Parts per million 3. Molarity of a solution 4. Molality of a solution 5. Mole fraction 6. Normality 7. Formality
O7. How would you express the concentration of a solution in percentage?
a. Mass percentage (w/w): The mass percentage of a component in a given solution is the mass of the component per hundred grams of the solution. = b. Volume percentage (V/V): The volume percentage is defined as the volume of the component per 100 parts by volume of the solution. Volume of the component 100 Total volume of solution c. Mass by volume percentage (w/V) : it is a Mass of solute dissolved in 100 ml of the solution. 2. Parts per million: When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as: Parts per million = Number of parts of the component ×106 Total number of parts of all components of the solution 3. Molarity of a solution: Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution, Molarity = Moles of solute Volume of solution in litre 4. Molality of a solution : Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as: Molality (m) = Moles of solute Mass of solvent in kg 5. Mole fraction : Commonly used symbol for mole fraction is x and subscript used on the right hand side of x denotes the component. Mole fraction is defined as the ratio of number of moles of one component to the total number of moles (solute and solvent) present in the solution. Mole fraction of a component = Number of moles of the component Total number of moles of all the components 6. Normality : It is the Number of gram equivalents of the solute dissolued per litre of the solution. It is denoted bv N. Normality (N) = Num!91of eramequivllents of solute Volume or "ototlotr-1., titr*.- _ Number"of gram equivalents of solute Volume of solution in mL normality are grn equivalent per ...(7) x 1000 7. Formality: It is the number of formula masses of the solute dissolved per litre of the solution. It is represented by F. Formality- _ Number of formula masses of solute Volume of the solution in litre.
O8. What is Solubility?
Solubility of a substance expresses the maximum amount of it which can be dissolved in a specific amount of solvent.
O9. Solubility of a Solid in a Liquid-
Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not. On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not. It is observed that polar solutes dissolve in polar solvents and non polar solutes in nonpolar solvents.
O10 . What are polar and non polar compounds? Explain with examples.
10. Polar compounds are those with distinct regions of positive and negative charge, as a result of bonding with atoms such as nitrogen, oxygen, or sulfur. Heavy oils generally contain greater proportions of higher boiling, more aromatic, and heteroatom-containing (N-, O-, S-, and metal-containing) constituents. Nonpolar compounds are compounds that do not have slightly negative and positive charges within the compound. The electronegativity differences of nonpolar compounds are between 0 and .2. HCl is a polar molecule as chlorine has a higher electronegativity than the hydrogen. Examples of homonuclear nonpolar molecules are oxygen (O2) and nitrogen (N2).
O11. What do you mean by like dissolves like?
11. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.
O12. What is Dissolution?
12. When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution.
O13. What is crystallisation?
13. Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation.
O14. What is state of dynamic equilibrium ?
14. A stage is reached when the two processes that is dissolution as crystallization occur at the same rate. Under such conditions, number of solute particles going into solution will be equal to the solute particles separating out and a state of dynamic equilibrium is reached.
Solute + Solvent ↔ Solution
At this stage the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure.
O15. What is saturated solution?
15. Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.
O16. What is unsaturated solution?
16. An unsaturated solution is one in which more solute can be dissolved at the same temperature.
O17. Which are factors on which solubility of a substance depend?
17. The solubility of a substance depends upon i. the nature of solute ii. nature of the solvent iii. temperature and iv. pressure
O18. Explain Effect of temperature on solubility.
18. The solubility of a solid in a liquid is significantly affected by temperature changes. if the solute dissolves with absorption of heat (endothermic process), the solubility increases with rise in , temperature. if the solute dissolves with evolution of heat (exothermic process), the solubility decreases with rise in temperature.
O19. Explain Effect of pressure on solubility.
19. Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.
SVT031202.001 SVT031102.001 (Test) Class 12 Chemistry Test 1. State Henry’s law and mention some important applications. 2. The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas? 3. What is meant by positive and negative deviations from Raoult's law and how is the sign of Δmix H related to positive and negative deviations from Raoult's law? 4. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute? 5. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane? 6. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it. 7. Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. 8. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K. 9. A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K. 10. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B. 11. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration? SVT031102.002 12. Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water (iv) methanol and acetone (v) acetonitrile (CH3CN) and acetone (C3H6O). The most important type of inter-molecular attractive interaction in the given pairs are given below: (i) Van der waals interactions (ii) Van der waals interactions (iii) Ion dipole interactions (iv) Hydrogen bonding (v) Dipole dipole interactions 14. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN. n-Octane is non polar and can dissolve non-polar solutes. It cannot dissolve polar (and ionic) solutes. Cyclohexane is non-polar. Hence, easily soluble in n-octane. Methanol and acetonitrile are polar and have very low solubility in n-octane. KCl is ionic compound and hence, insoluble in n-octane. The increasing order for solubility in n-octane is as follows: KCl < CH3OH < CH3CN < Cyclohexane 15. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol. (i) Phenol (having polar-OH group)-Partially soluble. (ii) Toluene (non-polar)-Insoluble. (iii) Formic acid (form hydrogen bonds with water molecules )-Highly soluble. (iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble. (v) Chloroform (non-polar)-Insoluble. (vi) Pentanol (having polar-OH)-Partially soluble. 16. If the density of some lake water is 1.25g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake. Mass of sodium ions =92 g Molar mass of sodium ions =23 g/mol Number of moles of sodium ions =92g//23g mol-1 =4 Mass of water =1 kg Molality =Mass of water (in kg)number of moles of sodium ions=4/1=4 m 17. If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution. 18. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN. 19. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10–3 m aqueous solution required for the above dose. 20. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. SVT031102.003 21. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly. Fluorine, being the most electronegative, has the strongest electron withdrawing inductive effect (-I effect). Consequently, trifluoroacetic acid is the strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes most while acetic acid ionizes least. More the number of ions produced, more is the depression in freezing point. Thus, the depression in freezing point is maximum for the trifluoroacetic acid and minimum for acetic acid. The depression in freezing points are in the order: Acetic acid < trichloroacetic acid < trifluoracetic acid. SVT</sub>

Available Video lessons
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Click Here4 Assignment formulae molar mass and P,N,E with table
Click Here5 Molar mass illustrations
Click Here 7 Mole fraction ethylene glycol
Click Here 9 Solubility, factors, transition temperature, solubility of gases in liquids, Henry law and dalton’s observations
Click Here10 (Henry’s law) limitations applications, Num. Henry law
Click Here11.26 Colligative properties, relative lowering of vapour pressure

Click Here12.26 Elevation boiling point

Click Here13.26 Relative lowering of vapor pressure

Click Here14.26 Osmosis osmotic pressure

MCQs
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1. From the below options, choose the correct example for gaseous solutions.
(a) Oxygen dissolved in water
(b) Camphor in nitrogen gas
(c) Carbon dioxide dissolved in water
(d) Hydrogen in palladium

(b)
A gaseous solution is a solution in which the solvent is a gas.

2. Which among the following is an example of a solid solution?
(a) Copper dissolved in gold
(b) Ethanol dissolved in water
(c) Glucose dissolved in water
(d) Sodium chloride dissolved in water

(a)

3. The value of Henry’s constant KH is _. (a) greater for gases with higher solubility. (b) greater for gases with lower solubility. (c) constant for all gases. (d) not related to the solubility of gases.

(b)

4. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to _. (a) low temperature (b) low atmospheric pressure (c) high atmospheric pressure (d) both low temperature and high atmospheric pressure

(b) Body temperature of human beings remains constant.

5. What is the mole fraction of ethylene glycol in a solution containing 20g by mass?
(a) 0.022
(b) 0.054
(c) 0.068
(d) 0.090

(c)
The mole fraction of ethylene glycol is calculated as shown below:
The molecular mass of ethylene glycol (C2H6O2) = 24 + 6 + 32 = 62 g/mol.
The mass of ethylene glycol in solution = 20g.
The mass of water = 100 – 20 = 80g.
Thus, number of moles of ethylene glycol = mass of ethylene glycol ÷ molecular mass of ethylene glycol
Moles of ethylene glycol = 20 ÷ 62 = 0.322 moles.
Similarly, the number of moles of water = 80 ÷ 18 = 4.444 moles.
The mole fraction of ethylene glycol = 0.322 ÷ (0.322 + 4.444) = 0.068
(so1Ref)- MCQ5

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6. The solubility of a substance in a solvent depends on
(a) Temperature
(b) Pressure
(c) Nature of solute and solvent
(d) All of the above

(d)

7. In how many ways can the concentration of a solution be expressed?
(a) 1
(b) 3
(c) 5
(d) 8

The concentration of a solution can be expressed : Mass %, Volume %, Mole fraction, Parts per million, Mass by volume percentage, Molarity, Molality and Normality.

8. A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is __. (a) saturated (b) supersaturated (c) unsaturated (d) concentrated

(b) If added substance dissolves, the solution is unsaturated. If it does not dissolve solution is saturated. If precipitation occurs solution is supersaturated.

9. Which of the following does not dissolve in benzene?
(a) Naphthalene
(b) Anthracene
(c) C6H12O6
(d) All of the above

(c)
Glucose does not dissolve in benzene.

10. Choose the ideal solution from the following.
(a) Carbon disulphide and acetone
(b) Phenol and Aniline
(c) Chloroform and Acetone
(d) Ethyl iodide and ethyl bromide

(d)

11. How does the solubility of gasses vary with pressure?
(a) Increases with pressure
(b) Decreases with pressure
(c) First increases and then decreases
(d) No effect

(a)
The solubility of gasses in liquids increases with increase in pressure.

12. How does the solubility of gasses in a liquid vary with increase in temperature?
(a) Increases with temperature
(b) Decreases with temperature
(c) First increases and then decreases
(d) No effect

(b)

13. The term homogenous mixtures signifies that (a) its composition is uniform throughout the mixture. (b) its properties are uniform throughout the mixture. (c) both composition and properties are uniform throughout the mixture. (d) neither composition nor properties are uniform throughout the mixture.

(c) In homogeneous mixtures composition and properties both are uniform throughout the mixture.

14. Atomic mass is equal to (a) number of electrons of an atom (b) sum of the numbers of electrons and protons of an atom (c) sum of the numbers of neutrons and protons of an atom (d) none of these

(b) sum of the numbers of electrons and protons of an atom

15. Henry’s law constant for molality of methane is benzene at 298 K is 4.27 × 10⁵ mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg is (a) 1.78 × 10-3 (b) 17.43 (c) 0.114 (d) 2.814

(a)

16. Among the following substances the lowest vapour pressure is exerted by (a) water (b) alcohol (c) ether (d) mercury

(d) mercury

17. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is -. (a) less than the rate of crystallisation. (b) greater than the rate of crystallisation. (c) equal to the rate of crystallisation. (d) zero

(c)

18. If 2 gm of NaOH is present is 200 ml of its solution, its molarity will be (a) 0.25 (b) 0.5 (c) 5 (d) 10

(a) 0.25

19. The atmospheric pollution is generally measured in the units of (a) mass percentage (b) volume percentage (c) volume fraction (d) ppm

(d) ppm

20. A 5% solution of cane-sugar (molecular weight = 342) is isotonic with 1% solution of substance A. The molecular weight of X is (a) 342 (b) 171.2 (c) 68.4 (d) 136.8

(c) 68.4

21. 234.2 gm of sugar syrup contains 34.2 gm of sugar. What is the molal concentration of the solution. (a) 0.1 (b) 0.5 (c) 5.5 (d) 55

(b) 0.5

22. Which law explained solubility of gasses in a liquid?
(a) Charles law
(b) Henry’s law
(c) Raoult’s law
(d) Boyle’s law

(b)

23. Which of the following fluoride is used as rat poison? (a) CaF2 (b) KF (c) NaF (d) MgF2

(c) Sodium fluoride is used as rat poison.

24. Most of the processes in our body occur in (a) solid solution (b) liquid solution (c) gaseous solution (d) colloidal solution

(b) Almost all the processes in our body occur in liquid solution.

25. Choose the correct example for a non-ideal solution?
(a) Benzene + Toluene
(b) Hexane + Heptane
(c) Chlorobenzene + Bromobenzene
(d) Ethanol + Hexane

(d)

26. Which condition holds for the ideal solution?
(a) Change is volume is zero
(b) Change in volume is non-zero
(c) Change is enthalpy is non-zero
(d) None of the above

(a)

27. Which condition holds for a non-ideal solution?
(a) Change is volume is zero
(b) Change in volume is non-zero
(c) Change is enthalpy is zero
(d) None of the above

(b)

28. The law which indicates the relation’-hip between solubility of a gas in liquid and pressure rs (a) Raoult’s law (b) Henry’s law (c) Lowering of vapour pressure (d) Van’t Hoff lawAnswer

(b) Henry’s law

29. Molarity of liquid HCl will be, if density of solution is 1.17 gm/cc (a) 36.5 (b) 32.05 (c) 18.25 (d) 42.10

(b)

30. 1 M, 2.5 litre NaOH solution is mixed with another 0.5 M, 3 litre NaOH solution. Then find out the molarity of resultant solution (a) 0.80 M (b) 1.0 M (c) 0.73 M (d) 0.50 M

(c)

31. Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y? (a) 20 (b) 50 (c) 100 (d) 200

(b) 50

32. What does Henry’s constant depend upon?
(a) Nature of gas
(b) Nature of solvent
(c) Temperature
(d) All of the above

(d)

33. How is Henry’s constant dependent on temperature?
(a) Directly proportional
(b) Inversely proportional
(c) Varies exponentially
(d) None of the above

(a)

34. Which of the following units is useful in relating concentration of solution with its vapour pressure?
(a) Mole fraction
(b) Parts per million
(c) Mass percentage
(d) Molality

(a) Mole fraction

35. Which of the following mixture is(are) called solution?
(i) water + ammonia
(ii) water + acetone
(iii) acetone + alcohol
(iv) hexane + water
(a) (i), (ii) and (iii)
(b) (i), (iii) and (iv)
(c) (i) and (iv)
(d) (ii) and (iii)

(a) Hexane is not water soluble, hence solution is not formed.

36. A plant cell shrinks when it is kept in a
(a) hypotonic solution
(b) hypertonic solution
(c) isotonic solution
(d) pure water

(b) hypertonic solution

37. Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon ________.
(a) Temperature
(b) Nature of solute
(c) Pressure
(d) Nature of solvent

Answer: (c)

38. The relative lowering in vapour pressure is proportional to the ratio of number of
(a) solute molecules to solvent molecules
(b) solvent molecules to solute molecules
(c) solute molecules to the total number of molecules in solution
(d) solvent molecules to the total number of molecules in solution

(c) solute molecules to the total number of molecules in solution

39. Which of the following is a quantitative description of the solution?
(a) Dilute
(b) Concentrated
(c) Saturated
(d) Molar

(d) Dilute, concentrated and saturated terms are qualitative methods of description of concentration of solution whereas molar or molarity is quantitative method.

40. When a solute is present in trace quantities the following expression is used (a) Gram per million (b) Milligram percent (c) Microgram percent (d) Parts per million

(d)

41. How much oxygen is dissolved in 100 rnL water at 298 K if partial pressure of oxygen is 0.5 atm and K = 1.4 × 10-3 mol/L/atm?
(a) 22.4 mg
(b) 22.4 g
(c) 2.24 g
(d) 2.24 mg

(d) 2.24 mg

42. The molarity of the solution containing 7.1 g of Na2SO4 in 100 ml of aqueous solution is
(a) 2 M
(b) 0.5 M
(c) 1 M
(d) 0.05 M

(b)

43. Partial pressure of a solution component is directly proportional to its mole fraction. This is known as
(a) Henry’s law
(b) Raoult’s law
(c) Distribution law
(d) Ostawald’s dilution law

(b) Raoult’s law

44. 3 moles of P and 2 moles of Q are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are 80 and 60 torr respectively?
(a) 80 torr
(b) 140 torr
(c) 72 torr
(d) 70 torr

(c) 72 torr

45. Sprinkling of salt helps in clearing the snow covered roads in hills. The phenomenon involved in the process is (a) lowering in vapour pressure of snow (b) depression in freezing point of snow (c) melting of ice due to increase in temperature by putting salt (d) increase in freezing point of snow

(b) depression in freezing point of snow

46. Which of the following solutions shows positive deviation from Raoult’s law?
(a) Acetone + Aniline
(b) Acetone + Ethanol
(c) Water + Nitric acid
(d) Chloroform + Benzene

(b) Acetone + Ethanol

47. The system that forms maximum boiling azetrope is
(a) Acetone-chloroform
(b) ethanol-acetone
(c) n-hexane-n-heptane
(d) carbon disulphide-acetone

(a) Acetone-chloroform

48. A solution containing 10.2 g glycerine per litre is isotonic with a 2% solution of glucose. What is the molecular mass of glycerine?
(a) 91.8 g
(b) 1198 g
(c) 83.9 g
(d) 890.3 g

(a) 91.8 g

49. What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10°C ?
(a) 496 g
(b) 297 g
(c) 310 g
(d) 426 g

(b) 297 g

50. The osmotic pressure of a solution can be increased by
(a) increasing the volume
(b) increasing the number of solute molecules
(c) decreasing the temperature
(d) removing semipermeable membrane

(b) increasing the number of solute molecules

51. For carrying reverse osmosis for desalination of water the material used for making semipermeable membrane is
(a) potassium nitrate
(b) parchment membrane
(c) cellulose acetate
(d) cell membrane

(c) cellulose acetate

52. What will be the degree of dissociation of 0.1 M Mg(NO3)2 solution if van’t Hoff factor is 2.74?
(a) 75%
(b) 87%
(c) 100%
(d) 92%

(b) 87%

53. At equilibrium the rate of dissociation of a solid solute in a volatile liquid solvent is
(a) less than the rate of crystallisation
(b) greater than the rate of crystallisation
(c) equal to the rate of crystallisation
(d) zero

(c) equal to the rate of crystallisation

54. H2S is a toxic gas used in qualitative analysis. If solubility of H2S in water at STP is 0.195 m. what is the value of KH?
(a) 0.0263 bar
(b) 69.16 bar
(c) 192 bar
(d) 282 bar

55. What is an unsaturated solution? a) A solution that contains the maximum amount of solute possible at a given temperature and pressure b) A solution that contains less solute than a saturated solution at the same temperature and pressure c) A solution that contains more solute than a saturated solution at the same temperature and pressure d) A solution that contains no solute at all

B

56. What happens to the solubility of a solute in a solvent as the temperature increases? a) It decreases b) It remains the same c) It increases d) It fluctuates randomly

C

57. An unsaturated solution can be made more concentrated by which of the following methods? a) Increasing the temperature b) Decreasing the pressure c) Adding more solute d) Removing solvent

C

58. An unsaturated solution can be made less concentrated by which of the following methods? a) Decreasing the temperature b) Increasing the pressure c) Removing solute d) Adding solvent

A

59. An unsaturated solution can be converted into a saturated solution by which of the following methods? a) Decreasing the temperature b) Increasing the pressure c) Adding more solute d) Removing solvent

C

60. An unsaturated solution can be converted into a supersaturated solution by which of the following methods? a) Increasing the temperature b) Decreasing the pressure c) Adding more solute d) Removing solvent

C

61. What is the solubility of a solute in a given solvent at a specific temperature and pressure called? a) Solubility constant b) Solubility limit c) Solubility point d) Solubility coefficient

A

62. How does the solubility of a solute in a solvent change as the pressure of the solution increases? a) It increases b) It decreases c) It remains the same d) It fluctuates randomly

A

63. What is the difference between a saturated solution and an unsaturated solution? a) A saturated solution contains more solute than an unsaturated solution at the same temperature and pressure b) A saturated solution contains less solute than an unsaturated solution at the same temperature and pressure c) A saturated solution contains the maximum amount of solute possible at a given temperature and pressure, while an unsaturated solution contains less than the maximum d) A saturated solution is a solid solution while an unsaturated solution is a liquid solution

C

64. How does the solubility of a solute in a solvent change as the temperature of the solution increases? a) It increases b) It decreases c) It remains the same d) It fluctuates randomly

C https://www.learncbse.in/ncert-exemplar-problems-class-12-chemistry-solution/ https://ncert.nic.in/pdf/publication/exemplarproblem/classXII/chemistry/leep502.pdf 1. NE1 1. Which of the following units is useful in relating concentration of solution with its vapour pressure? (i) mole fraction (ii) parts per million (iii) mass percentage (iv) molality (a) Mole fraction is useful in’relating vapour pressure with concentration of solution. According to Raoult’s law, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. A is one component. 2. NE2 On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid? (i) Sugar crystals in cold water. (ii) Sugar crystals in hot water. (iii) Powdered sugar in cold water. (iv) Powdered sugar in hot water. (d) Since the solution is cool to touch, the dissolution is endothermic. Therefore, high temperature will favour dissolution. Further, powdered sugar has large surface area and is favourable for dissolution. 3. NE3 At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________. (i) less than the rate of crystallisation (ii) greater than the rate of crystallisation (iii) equal to the rate of crystallisation (iv) zero (c) At equilibrium the rate of dissolution of solid solute is equal to rate of crystallisation. 4. NE4 A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is _________. (i) saturated (ii) supersaturated (iii) unsaturated (iv) concentrated (b) When small amount of solute is added to its solution and it does not dissolve and get precipitated then this solution is supersaturated solution. 5. NE5 Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon ____________. (i) Temperature (ii) Nature of solute (iii) Pressure (iv) Nature of solvent (c) Solubility of a solid in liquid does not depend on pressure because solid is practically incompressible. 6. NE6 Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________. (i) low temperature (ii) low atmospheric pressure (iii) high atmospheric pressure (iv) both low temperature and high atmospheric pressure 7. NE7 Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law? (i) Methanol and acetone. (ii) Chloroform and acetone. (iii) Nitric acid and water. (iv) Phenol and aniline. 8. NE8 Colligative properties depend on ____________. (i) the nature of the solute particles dissolved in solution. (ii) the number of solute particles in solution. (iii) the physical properties of the solute particles dissolved in solution. (iv) the nature of solvent particles. 9. NE9 Which of the following aqueous solutions should have the highest boiling point? (i) 1.0 M NaOH (ii) 1.0 M Na2SO4 (iii) 1.0 M NH4NO3 (iv) 1.0 M KNO3 10. NE10 The unit of ebulioscopic constant is _______________. (i) K kg mol–1 or K (molality)–1 (ii) mol kg K–1 or K–1(molality) (iii) kg mol–1 K–1 or K–1(molality)–1 (iv) K mol kg–1 or K (molality)