Chapter 2 : SOLUTIONS
ONE MARK QUESTIONS
v OBJECTIVE TYPE MCQs
1.) The molarity of the solution containing 7.1 g of Na2SO4 in 100 ml of aqueous solution is
(a) 2M
(b) 0.5M
(c) 1M
(d) 0.05M
Ans: Molarity = no. of moles/ Vol. of solution(L) = given mass x 1000 / molar mass x vol.(ml)
= 7.1 x 1000 /142 x 100 = 0.5M
2.) Which of the following statements is incorrect?
(a) A solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.
(b) An unsaturated solution is one in which more solute can be dissolved at the same temperature.
(c) The solution which is in dynamic equilibrium with undissolved solute is the saturated solution.
(d) The minimum amount of solute dissolved in a given amount of solvent is its solubility
Ans: The maximum amount of solute dissolved in a given amount of solvent is its solubility.
3.) Which is an application of Henry’s law?
(a) Spray paint
(b) Bottled water
(c) Filling up a tyre
(d) Soft drinks (soda)
Ans: To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure
(a) Water - Nitric acid
(b) Benzene - Methanol
(c) Water - Hydrochloric acid
(d) Acetone – Chloroform
Ans: Positive deviations are shown by such solutions in which solvent-solvent and solute-solute interactions are stronger than the solute-solvent interactions. In such solution, the interactions among molecules becomes weaker. Therefore, their escaping tendency increases which results in the increase in their partial vapour pressures. In pure methanol there exists intermolecular H–bonding.
On adding benzene, its molecules come between ethanol molecules there by breaking H-bonds which weaken intermolecular forces. This results in increase in vapour pressure.
(a) solute molecules to the solvent molecules
(b) solute molecules to the total molecules in the solution
(c) solvent molecules to the total molecules in the solution
(d) solvent molecules to the total number of ions of the solute.
Ans: According to Raoult's law, the relative lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute present in the solution.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
1.) Assertion: Molarity of a solution in liquid state changes with temperature.
Reason: The volume of a solution changes with change in temperature.
Ans: (a)
2.) Assertion: If a liquid solute more volatile than the solvent is added to the solvent, the vapour pressure of the solution may increase i.e., ps > po.
Reason: In the presence of a more volatile liquid solute, only the solute will form the vapours and solvent will not.
Ans: (c)
3.) Assertion: If one component of a solution obeys Raoult’s law over a certain range of composition, the other component will not obey Henry’s law in that range.
Reason: Raoult’s law is a special case of Henry’s law
Ans: (b)
4.) Assertion: When methyl alcohol is added to water, boiling point of water increases.
Reason: When a volatile solute is added to a volatile solvent elevation in boiling point is observed.
Ans: (d)
Reason: The lowering of vapour pressure of a solution causes depression in the freezing point
Ans: (a)
6. Assertion: Molality is a better method to express concentration than molarity
Reason: Molality is defined in terms of mass of solvent and not mass of solution.
7. Assertion: Soda bottles are sealed under high pressure.
Reason: High pressure increases the solubility of carbon dioxide gas.
Ans- a
8. Assertion: There will not be any change in concentration of an ethanol and water mixture containing 85% ethanol by volume on boiling.
Reason: It is because azeotropes boil out in constant composition.
Ans- d
9. Assertion: Benzene and hexane form an ideal solution.
Reason: Both benzene and hexane are hydrocarbons.
Ans- c
10. Assertion: 1 molar NaCl solution has higher boiling point than one molar urea.
Reason: NaCl dissociates into ions in solution.
Ans- a
11. Assertion: A raw mango placed in concentrated salt solution loses water shrivel into pickle.
Reason: The salt solution is hypotonic compared to the raw mango.
Ans- c
12. Assertion: Helium is mixed with nitrogen and oxygen in diving cylinders
Reason: Helium has comparatively low KH(Henry constant) value.
Ans- c
13. Assertion: Ice containing dissolved sugar will melt at a higher temperature than pure ice.
Reason: Dissolving sugar in water leads to a depression in freezing point.
Ans-d
14. Assertion: Molar mass of acetic acid in benzene calculated using colligative property is almost double the actual value.
Reason: Acetic acid dimerises in solution.
Ans- a
15. Assertion: Vapour pressure of a solution is more that of the pure solvent.
v CASE BASED QUESTIONS
Read the following passage and answer the questions that follow:
Solutions are homogeneous mixture of two or more substances. Ideal solution follow Raoult’s law. The vapour pressure of each component is directly proportional to their mole fraction if both solute and solvent are volatile. The relative lowering of vapour pressure is equal to mole fraction of solute if only solvent is volatile. Non-ideal solution form azeotropes which cannot be separated by fractional distillation. Henry’s law is special case of Raoult’s law applicable to gases dissolved in liquids. Colligative properties depend upon number of particles of solute. Relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure are colligative properties which depend upon mole fraction of solute, molality and molarity of solutions.
(1) 50 ml of an aqueous solution of glucose (Molar mass 180 g/mol) contains 6.02 × 1022 molecules.
What is its molarity?
Ans. 6.02 x 1023 molecules → 1 mole of glucose
6.02 x 1022 molecules → 0.1 moles
Molarity = 0.1/0.05 = 2 M
2) Identify which liquid has lower vapour pressure at 90°C if boiling point of liquid ‘A’ and ‘B’ are 140°C and 180° respectively.
Ans. ‘B’ will have lower vapour pressure because its boiling point is higher.
(3) What type of azeotropes are formed by non-ideal solution showing negative deviation from Raoult’s law?
Ans. Maximum boiling azeotropes.
(4) Why meat is preserved for longer time by salting?
Ans. Salt inhibits the growth of microorganisms by drawing out water from microbial cells through osmosis 20% salt is needed to kill most species of unwanted bacteria.
(5) For a 5% solution of area (molar mass 60 g/mol), calculate the osmotic pressure at 300 K (R= 0.0821 L atm k–1).
Ans. ∏v = nRT
∏ × 0.1 L = (5/60) × 0.0821 × 300 = 20.52 atm
Ans: Henry’s law states that “The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution”
2. Define an ideal solution and write one of its characteristics.
Ans: Those solutions which obey Raoult’s law at all concentrations are called ideal solutions.
Characteristic of an ideal solution:
There will be no change in enthalpy ∆Hmix = 0 and
volume change on mixing ∆Vmix = 0,
3. Define osmotic pressure of a solution.
Ans: It is the external pressure which is applied on the solution side which is sufficient to prevent the entry of the solvent through semi-permeable membrane.
Ans: All those properties which depend on the number of solute particles in solution irrespective of the nature of solute are called as colligative properties.
5. Give an example of gaseous solution and liquid solution each.
Ans: Gaseous solution- Water vapour (Liquid in gas)
Liquid solution- Ethanol dissolved in water (Liquid in liquid)
1.) 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.
Ans- Solubility of H2S gas = 0.195 m
= 0.195 mole in 1 kg of solvent 1kg of solvent = 1000g
2.) Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that moles of urea = 0.25 mole mass of solvent (NH2CONH2) = 60 g mol-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g
3.) Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: Mass of solution = Mass of C6H6 + Mass of CCl4
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl4 = 122/144 x 100 = 84.72 %
Ans: 30% by mass of C6H6 in CCl4 => 30 g C6H6 in 100 g solution
.’. no. of moles of C6H6(nC6h6) = 30/78 = 0.385
5.) 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. (Kf for CH3COOH) = 3·9 K kg mol-1)
Ans:
6. Differentiate between a minimum boiling azeotropes and max azeotropes
7. describe type of solutions as isotonic, Hypertonic and hypotonic solution.
8. Two application of Henry's law.
ans- 1.dissolving CO2 in soda water/ cold drinks
2. to avoids bends
3. anoxia
THREE MARKS QUESTIONS
1.) Calculate the mass of a non-volatile solute(molecular mass 40 g mol-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%.
Ans: According to Raoult’s Law,
2.) Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20g 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.
Ans:
3) 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 35.0 g of octane?
Ans: