CBSE Class 12 Chemistry Solutions Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Two liquids A and B on mixing produce a warm solution. Which type of deviation from Raoult's law does it show?
Answer:
Warming up of the solution means that the process of mixing is exothermic, i.e., ΔH_mixing = −ve. This implies that the solution shows a negative deviation.

Teacher's Note:
1. Heat released means stronger A–B attractions formed, which is the mark of negative deviation.

Question 2. Why does a solution of ethanol and cyclohexane show positive deviation from Raoult's law?
Answer:
On adding cyclohexane, its molecules get inbetween the molecules of ethanol thus breaking the hydrogen bonds present between ethanol molecules and reducing ethanol-ethanol interactions.

Teacher's Note:
1. Weaker interactions after mixing let molecules escape more easily, so the vapour pressure is higher than expected.

Question 3. After removing the outer shell of two eggs in dil. HCl, one is placed in distilled water and the other is placed in a saturated solution of NaCl. What will you observe and why?
or
A peeled egg swells when dipped in water while shrinks in saturated brine solution. Why? [Uttarakhand Board 2012]
Answer:
Egg in water will swell while egg in NaCl solution will shrink. This is because as a result of osmosis, the net flow of solvent is from less concentrated to more concentrated solution.

Teacher's Note:
1. The egg membrane acts as the semipermeable membrane; water moves towards the more concentrated side.

Question 4. What will happen to the boiling point of a solution if the weight of the solute dissolved is doubled but the weight of solvent taken is halved?
Answer:
Elevation in boiling point will be four times because ΔT_b = \( \frac{1000\, K_b\, w_2}{M_2\, w_1} \).

Teacher's Note:
1. ΔT_b ∝ w₂/w₁: doubling the numerator and halving the denominator multiplies it by 4.

Question 5. Why is camphor preferred as a solvent in finding the molecular mass of naphthalene by Rast method?
or Why is camphor preferred in the determination of ΔT_f?
Answer:
Camphor has a large value of K_f, viz, 39.7°C so that depression in melting point is large for a solution of naphthalene in camphor.

Teacher's Note:
1. A large K_f gives a large, easily measured ΔT_f, so an ordinary thermometer is enough.

Question 6. Why boiling point of water is increased on addition of sodium chloride into it?
Answer:
When NaCl (or any other non-volatile solute) is dissolved in water, vapour pressure of water decreases (because some of the solvent molecules on the surface are replaced by the molecules of the solute which are non-volatile). The solution has to be heated more to make vapour pressure equal to the external pressure. Hence, boiling point is increased.

Teacher's Note:
1. Boiling point is the temperature at which vapour pressure equals external pressure – always start the reasoning there.

Question 7. What freezes out first when a solution of common salt is cooled?
Answer:
Water as ice.

Teacher's Note:
1. On cooling a dilute solution, pure solvent separates as solid at the freezing point of the solution.

Question 8. Why melting point of a substance is used as a criterion for testing the purity of the substance.
Answer:
A pure compound has a sharp melting point. Impurities present, if any, lower the melting point of the compound (just as depression in freezing point takes place).

Teacher's Note:
1. An impure sample also melts over a range of temperature rather than sharply.

Question 9. How is it that alcohol (ethanol) and water are miscible in all proportions ? (CBSE 2007)
Answer:
(i) Both are polar and hence miscible.
(ii) They form hydrogen bonds with each other. This further increases their miscibility.

Teacher's Note:
1. "Like dissolves like": similar polarity plus solute-solvent hydrogen bonding makes the mixing complete.

Question 10. Give an example of a solution containing a liquid solute in a solid solvent.
Answer:
Hydrated salts like CuSO₄.5H₂O.

Teacher's Note:
1. Amalgam of mercury with sodium is another standard example of a liquid (Hg) dissolved in a solid (Na).

Question 11. Give one example each of solid in gas and liquid in gas solutions.
Answer:
Iodine vapour in air, humidity in air.

Teacher's Note:
1. In gaseous solutions the solvent is a gas; camphor in nitrogen gas is another solid-in-gas example.

Question 12. Define ‘molality’. (CBSE 2017)
Answer:
Molality of a solution is the moles of the solute dissolved in 1 kg of the solvent.

Teacher's Note:
1. Note: 1 kg of solvent, not of solution; this is the most common slip in this definition.

Question 13. What do you mean by saying that the molality of a solution is 0.1 ?
Answer:
It means that 0.1 mol of the solute is dissolved in 1 kg of the solvent.

Teacher's Note:
1. Molality is written with the symbol m and unit mol kg⁻¹.

Question 14. What is the relation between normality and molarity of a given solution of H₂SO₄ ?
Answer:
Normality = 2 × Molarity.

Teacher's Note:
1. Normality = Molarity × basicity for acids; H₂SO₄ is dibasic, so the factor is 2.

Short Answer Questions (2 Marks)

Question 15. Why is it advised to add ethylene glycol to water in a car radiator in hill station?
Answer:
Ethylene glycol acts as anti-freeze: it lowers the freezing point of water, so the water in the radiator does not freeze in cold weather.

Teacher's Note:
1. Ethylene glycol is used because it is non-volatile and highly soluble in water.

Question 16. Calculate the molarity of pure water (d = 1 g mL⁻¹).
Answer:
Density of water = 1 g mL⁻¹
Mass of 1000 mL of water = V × d = 1000 mL × 1 g mL⁻¹ = 1000 g
Moles of water = \( \frac{1000}{18} \) = 55.55 mol
Now, 55.55 mol of H₂O are present in 1000 mL or 1 L of water.
So, molarity = 55.55 M

Teacher's Note:
1. The same number, 55.5 mol per kg, is used when converting molality into mole fraction for dilute aqueous solutions.

Question 17. Define Henry’s law. Give their two application.
Answer:
Henry’s law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. In terms of mole fraction, p = K_Hx.
Applications:
(i) To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.
(ii) Scuba divers use air diluted with helium to avoid the bends caused by nitrogen dissolving in the blood at high pressure.

Teacher's Note:
1. Another application: at high altitudes the low partial pressure of oxygen lowers its concentration in blood, causing anoxia.

Question 18. The dissolution of ammonium chloride in water is endothermic process. What is the effect of temperature on its solubility?
Answer:
Since dissolution of NH₄Cl in water is endothermic process, its solubility increases with rise in temperature (i.e., Le-Chatelier principle).

Teacher's Note:
1. For exothermic dissolution the opposite holds: solubility falls as temperature rises.

Question 19. Two liquids A and B boil at 145°C and 190°C respectively. Which of them has higher vapour pressure at 80°C?
Answer:
Lower the boiling point more volatile is the respective compound. Therefore, liquid A will have higher vapour pressure at 80°C.

Teacher's Note:
1. A lower boiling point means weaker intermolecular forces and a higher vapour pressure at any given temperature.

Question 20. Why is liquid ammonia bottle first cooled in ice before opening it?
Answer:
At room temperature, the vapour pressure of liquid ammonia is very high. On cooling vapour pressure decreases, therefore the liquid ammonia will not splash out.

Teacher's Note:
1. Vapour pressure always rises with temperature, so cooling is the simple way to reduce it.

Question 21. Which colligative property is preferred for the molar mass determination of macromolecules?
Answer:
Osmotic pressure measurement is preferred for molar mass determination because:
(a) even in dilute solution the osmotic pressure values are appreciably high and can be measured accurately.
(b) osmotic pressure can be measured at room temperature.

Teacher's Note:
1. Measuring at room temperature matters because many biomolecules are unstable at higher temperatures.

Question 22. Which solution has higher concentration, 1 molar or 1 molal solution of the same solute? Give reason.
Answer:
In aqueous solution, 1M solution means 1 mole of the solute in 1000 cc of the solution whereas 1 m solution means 1 mole of the solute in 1000 g of water (= 1000 cc of water). Total volume of 1 m solution > 1000 cc due to presence of extra 1 mole of the solute. Hence, number of moles/cc in 1 m solution will be less than that in 1 M solution. In other words, 1 M is more concentrated than 1 m.
In non-aqueous solution, 1 M can be greater than, less than or equal to 1 m, depending upon the density of the solution.

Teacher's Note:
1. Molarity is per litre of solution, molality per kg of solvent – picture where the solute's own volume goes.

Question 23. V₁ cc of solution having molarity M₁ is diluted to have molarity M₂. Derive expression (in terms of M₁, M₂ and V₁) for the volume of water required to be added.
Answer:
Suppose the final volume after dilution is V₂
Then M₁V₁ = M₂V₂ or V₂ = \( \frac{M_1 V_1}{M_2} \)
∴ Volume of water required to be added = V₂ − V₁ = \( \frac{M_1 V_1}{M_2} \) − V₁
= \( \left(\frac{M_1}{M_2} - 1\right) V_1 \) = \( \left(\frac{M_1 - M_2}{M_2}\right) V_1 \)

Teacher's Note:
1. On dilution the moles of solute stay the same, which is the whole basis of M₁V₁ = M₂V₂.

Question 24. Why the solubility of Glauber's salt (Na₂SO₄.10H₂O) first increases upto 32.4°C and then decreases?
Answer:
Upto 32.4°C, Glauber's salt remains in the hydrated form whose dissolution is endothermic. Hence, its solubility increases with temperature. Beyond 32.4°C, it changes into anhydrous salt whose dissolution is exothermic. Hence, solubility decreases with increase of temperature.

Teacher's Note:
1. The temperature at which the trend reverses (32.4°C here) is called the transition temperature.

Question 25. Give reasons for the following:
(a) Aquatic species are more comfortable in cold water than warm water.
(b) At higher altitudes, people suffer from anoxia resulting in inability to think. [CBSE 2019]
Answer:
(a) Dissolution of oxygen in water is exothermic, so its solubility decreases with rise in temperature (K_H increases). Cold water therefore contains more dissolved oxygen than warm water, and aquatic species are more comfortable in it.
(b) At high altitude, partial pressure of oxygen in the air is low. As a result there is low concentration of oxygen in the blood and tissues.

Teacher's Note:
1. Both parts are applications of Henry's law: (a) through temperature, (b) through partial pressure.

Question 26. Derive the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid.
Answer:
According to Raoult's law, \( \frac{p^\circ - p_s}{p^\circ} \) = x₂ (mole fraction of the non-volatile solute)
Mole fraction of the volatile liquid, x₁ = 1 − x₂ or x₂ = 1 − x₁
∴ \( \frac{p^\circ - p_s}{p^\circ} \) = 1 − x₁ or x₁ = 1 − \( \frac{p^\circ - p_s}{p^\circ} \) = \( \frac{p_s}{p^\circ} \)

Teacher's Note:
1. The final result p_s = p°x₁ is just Raoult's law for the solvent – a quick check of the derivation.

Question 27. Derive a relationship between mole fraction and vapour pressure of a component of an ideal solution in the liquid phase and vapour phase.
Answer:
Consider an ideal solution of two volatile liquids A and B with mole fractions x_A and x_B in the liquid phase.
By Raoult's law, p_A = p_A° x_A and p_B = p_B° x_B; P_total = p_A° x_A + p_B° x_B
If y_A and y_B are the mole fractions in the vapour phase, then by Dalton's law of partial pressures, p_A = y_A P_total and p_B = y_B P_total
∴ y_A = \( \frac{p_A^\circ x_A}{p_A^\circ x_A + p_B^\circ x_B} \) and y_B = \( \frac{p_B^\circ x_B}{p_A^\circ x_A + p_B^\circ x_B} \)

Teacher's Note:
1. Raoult's law links liquid mole fraction to partial pressure; Dalton's law links partial pressure to vapour mole fraction – combine the two.

Short Answer Questions (3 Marks)

Question 28. Discuss the effect of temperature on the solubility of solids in liquids.
Answer:
The effect of temperature on solubility depends upon the heat of solution, according to Le Chatelier’s principle.
(i) If dissolution is endothermic (Δ_solH > 0), solubility increases with rise in temperature, e.g., KNO₃, NH₄Cl, sugar.
(ii) If dissolution is exothermic (Δ_solH < 0), solubility decreases with rise in temperature, e.g., Ce₂(SO₄)₃, Li₂CO₃.
(iii) For some substances like NaCl, the solubility changes only slightly with temperature.
(iv) Some substances show irregular behaviour, e.g., solubility of Na₂SO₄.10H₂O first increases upto 32.4°C (transition temperature) and then decreases.

Teacher's Note:
1. Pressure has no significant effect on solubility of solids in liquids because solids and liquids are almost incompressible.

Question 29. State Raoult’s law. Derive its mathematical expression for a solution of a non-volatile solute in a volatile solvent. (J & K Board 2011, MP Board 2012)
Answer:
Raoult’s law : For a solution of a non-volatile solute, the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.
Let n₁ moles of solvent and n₂ moles of non-volatile solute be present. As only the solvent contributes to the vapour pressure,
p_s ∝ x₁ or p_s = p° x₁ (when x₁ = 1, p_s = p°, the V.P. of pure solvent)
But x₁ = 1 − x₂ ∴ p_s = p°(1 − x₂) or p° − p_s = p°x₂
or \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} = x_2 \)
i.e., the relative lowering of vapour pressure is equal to the mole fraction of the solute.

Teacher's Note:
1. Start from p_s = p°x₁ and replace x₁ by 1 − x₂; that single step is the whole derivation.

Question 30. What do you understand by Relative lowering of vapour pressure? How is it used to determine molecular mass of the solute? (Uttarakhand Board 2012)
Answer:
The ratio of lowering of vapour pressure to the vapour pressure of the pure solvent, (p° − p_s)/p°, is called relative lowering of vapour pressure. By Raoult’s law it is equal to the mole fraction of the solute.
Determination of molecular mass : If w₂ g of solute of molar mass M₂ is dissolved in w₁ g of solvent of molar mass M₁, then for a dilute solution (n₂ << n₁),
\( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_2 M_1}{M_2 w_1} \)
∴ \( M_2 = \frac{w_2 \times M_1 \times p^\circ}{w_1 \times (p^\circ - p_s)} \)
Thus, by measuring p° and p_s, the molar mass M₂ of the solute can be calculated.

Teacher's Note:
1. The approximation n₁ + n₂ ≈ n₁ is valid only for dilute solutions; use the full form if the solution is concentrated.

Question 31. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused ? Explain with one example of each type. (CBSE 2010)
Answer:
Positive deviation : When the vapour pressure of the solution is higher than that predicted by Raoult’s law, the solution shows positive deviation. It is caused when the A–B interactions are weaker than A–A and B–B interactions, so molecules escape more easily. Here Δ_mixH > 0 and Δ_mixV > 0. Example : ethanol + water, ethanol + acetone (hydrogen bonds of ethanol are broken).
Negative deviation : When the vapour pressure of the solution is lower than that predicted by Raoult’s law, the solution shows negative deviation. It is caused when the A–B interactions are stronger than A–A and B–B interactions. Here Δ_mixH < 0 and Δ_mixV < 0. Example : chloroform + acetone (a hydrogen bond is formed between CHCl₃ and acetone), HNO₃ + water.

Teacher's Note:
1. Link each deviation to its signs: positive deviation, ΔH and ΔV both positive; negative deviation, both negative.

Teacher's Note:
1. Point 5 links deviations to azeotropes: higher vapour pressure means a lower boiling point.

Teacher's Note:
1. Always quote the ΔH and ΔV conditions; they are the quickest scoring differences.

Question 34. What is Raoult’s law ? State the factors responsible for deviations from this law. Illustrate deviations with suitable examples.
Answer:
Raoult’s law : For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution (p_A = p_A°x_A).
Factors responsible for deviations : (i) Difference in the strength of A–B interactions compared to A–A and B–B interactions. (ii) Breaking of hydrogen bonds or association on mixing (e.g. ethanol molecules lose hydrogen bonds when mixed with acetone or cyclohexane). (iii) Formation of new hydrogen bonds or complexes between the components (e.g. CHCl₃ and acetone). (iv) Dissociation of a component in the solution.
Positive deviation : A–B attractions weaker, vapour pressure higher than expected; e.g. ethanol + acetone, carbon disulphide + acetone.
Negative deviation : A–B attractions stronger, vapour pressure lower than expected; e.g. chloroform + acetone, phenol + aniline, nitric acid + water.

Teacher's Note:
1. Name the interaction change (broken or new hydrogen bonds) when giving an example; that is the reasoning examiners look for.

Question 35. What are ideal and non-ideal solutions ? Give reasons for the formation of such solutions. Give one example in each case. (Raj. Board 2011, Karnataka Board 2012)
Answer:
Ideal solution : A solution which obeys Raoult’s law over the entire range of concentration, with Δ_mixH = 0 and Δ_mixV = 0. It is formed when the two components have similar structures and polarities so that A–B interactions are nearly the same as A–A and B–B interactions. Example : n-hexane + n-heptane, benzene + toluene.
Non-ideal solution : A solution which does not obey Raoult’s law, with Δ_mixH ≠ 0 and Δ_mixV ≠ 0. It is formed when A–B interactions are different (weaker or stronger) from A–A and B–B interactions. Example : ethanol + water (positive deviation), chloroform + acetone (negative deviation).

Teacher's Note:
1. Similar size, structure and polarity of the two liquids is the condition for an ideal solution.

Question 36. What are non-ideal solutions? What are their different types? Explain giving examples. (Chhatisgarh Board 2012)
Answer:
Solutions which do not obey Raoult’s law over the entire range of concentration are called non-ideal solutions. For them Δ_mixH ≠ 0 and Δ_mixV ≠ 0.
They are of two types :
(i) Showing positive deviation : The vapour pressure is higher than that predicted by Raoult’s law because A–B interactions are weaker than A–A and B–B interactions; Δ_mixH > 0, Δ_mixV > 0. Examples: ethanol + water, ethanol + acetone, CS₂ + acetone.
(ii) Showing negative deviation : The vapour pressure is lower than predicted because A–B interactions are stronger; Δ_mixH < 0, Δ_mixV < 0. Examples: chloroform + acetone, phenol + aniline, HNO₃ + water.

Teacher's Note:
1. Remember one example of each type that you can explain in terms of hydrogen bonding.

Long Answer Questions (5 Marks)

Question 37. (a) Define Raoult’s law of binary solution containing non-volatile solute in it.
(b) On dissolving 3.24 g of sulphur in 40 g of benzene, boiling point of solution was higher than that of benzene by 0.81 K (K_b = 2.53 K kg mol⁻¹). What is molecular formula of sulphur? (Atomic mass S = 32 g mol⁻¹)
Answer:
(a) At a given temperature, the vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.
(b) M_B = \( \frac{K_b \times W_B \times 1000}{\Delta T_b \times W_A} \) = \( \frac{2.53 \times 3.24 \times 10^{3}}{0.81 \times 40} \) = 253 g mol⁻¹
Let the molecular formula of sulphur = S_x
Atomic mass of sulphur = 32; Molecular mass = 32 × x
32x = 253; x = 7.91 ≈ 8
Molecular formula of sulphur = S₈

Teacher's Note:
1. Find the molar mass from the colligative property first, then divide by the atomic mass to get the atomicity.

Question 38. (a) Outer shells of two eggs are removed. One of the egg is placed in pure water and the other is placed in saturated solution of NaCl. What will be observed and why?
(b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is a non-electrolyte, determine the molar mass.
Answer:
(a) In pure water the egg swells and in saturated solution of NaCl it will shrink. This is due to osmosis: water enters the egg through its membrane from pure water, and water comes out of the egg into the more concentrated NaCl solution.
(b) Mass of gene fragment = 8.95 mg = 8.95 × 10⁻³ g
Volume of water = 35.0 mL = 35 × 10⁻³ L
π = 0.335 torr = \( \frac{0.335}{760} \) atm
Temp = 25 + 273 = 298 K
π = \( \frac{W_B RT}{M_B \times V} \)
\( \frac{0.335}{760} \) = \( \frac{8.95 \times 10^{-3} \times 0.0821 \times 298}{M_B \times 35 \times 10^{-3}} \)
M_B = 14193 g mol⁻¹ (≈ 1.42 × 10⁴ g mol⁻¹)

Teacher's Note:
1. Convert torr to atm (divide by 760) when using R = 0.0821 L atm K⁻¹ mol⁻¹.

Question 39. (a) Define van’t Hoff factor.
(b) Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na₂SO₄. If this solution actually freezes at −0.320°C, what would be the value of van’t Hoff factor? (K_f = 1.86°C kg mol⁻¹)
Answer:
(a) Van’t Hoff factor: It is the ratio of the normal molar mass to the observed molar mass of the solute.
(b) ΔT_f = K_f × m
ΔT_f = 1.86 × 0.0711 = 0.132°C (expected)
Observed freezing point depression = 0 − (−0.320) = 0.320°C
i = \( \frac{\text{Observed freezing point depression}}{\text{Calculated freezing point depression}} \)
= \( \frac{0.320}{0.132} \) = 2.42

Teacher's Note:
1. i comes out below the ideal value of 3 for Na₂SO₄ because of interionic attractions in the solution.

Question 40. (a) What is the value of i when solute is associated and dissociated?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr₂ in 200 g of water. (Molar mass of MgBr₂ = 184, K_f = 1.86 K kg mol⁻¹)
Answer:
(a) i < 1 when solute is associated and i > 1 when solute is dissociated.
(b) m = \( \frac{n_B \times 1000}{W_A\,(g)} \) = \( \frac{W_B \times 1000}{M_B \times W_A} \) = \( \frac{10.50 \times 1000}{184 \times 200} \) = 0.2853 m
MgBr₂ ionizes as MgBr₂ → Mg²⁺ + 2Br⁻; i = 3
ΔT_f = i × K_f × m = 3 × 1.86 × 0.2853 = 1.59
Freezing point = 0 − 1.59°C = −1.59°C

Teacher's Note:
1. The freezing point of the solution is 0 − ΔT_f; do not report ΔT_f itself as the freezing point.

Question 41. (a) What is the value of i for Al₂(SO₄)₃ when it is completely dissociated?
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250 g of water. (K_b = 0.512 K kg mol⁻¹ and molar mass of NaCl = 58.44 g mol⁻¹)
Answer:
(a) Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻; i = 5
(b) ΔT_b = \( \frac{iK_b \times 1000 \times W_B}{W_A \times M_B} \)
NaCl → Na⁺ + Cl⁻; i = 2
ΔT_b = \( \frac{2 \times 0.512 \times 1000 \times 15}{250 \times 58.44} \) = 1.05
Boiling point of solution = 100 + 1.05 = 101.05°C

Teacher's Note:
1. Count all ions: 2 Al³⁺ + 3 SO₄²⁻ = 5 particles per formula unit.