CBSE Class 12 Chemistry Solutions Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Define mole fraction. [CBSE Delhi 2012]
Answer:
Mole fraction of a component in a solution may be defined as the ratio of moles of that component to the total number of moles of all the components present in the solution.

Teacher's Note:
1. The mole fractions of all components of a solution always add up to 1 – a quick check in numericals.

Question 2. How does the molarity of a solution change with temperature?
Answer:
Molarity decreases with increase in temperature as volume of solution increases with increase in temperature.

Teacher's Note:
1. Molality and mole fraction are independent of temperature because they involve only masses.

Question 3. When and why is molality preferred over molarity in handling solutions in chemistry?
Answer:
Molality is preferred in studies that involves changes in temperature as in some of the colligative properties of solutions. This is because molality depends on masses of solvent which do not change with temperature.

Teacher's Note:
1. That is why ΔT_b and ΔT_f formulas use molality, while osmotic pressure (measured at a fixed temperature) uses molarity.

Question 4. The dissolution of ammonium chloride in water is an endothermic process but still it dissolves in water readily. Why? [HOTS]
Answer:
This is because of entropy change. In this case, ΔS is +ve.
NH₄Cl(aq) → NH₄⁺(aq) + Cl⁻(aq)
The ions that were held together in crystalline solid are free and moving in all possible directions. Its entropy has increased and this makes TΔS > ΔH, i.e., ΔG = −ve.

Teacher's Note:
1. Use ΔG = ΔH − TΔS: a positive ΔH can still give a negative ΔG if the entropy gain is large enough.

Question 5. Why are the aquatic species more comfortable in cold water in comparison to warm water? [CBSE 2019 (56/2/3), (56/4/1)]
Answer:
At a given pressure, the solubility of oxygen in water increases with decrease in temperature. Presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

Teacher's Note:
1. K_H of a gas increases with temperature, so gas solubility falls as water gets warmer.

Question 6. What is the similarity between Raoult’s law and Henry’s law? [CBSE Delhi 2014]
Answer:
The similarity between Raoult’s law and Henry’s law is that in both the laws, the partial vapour pressure of the volatile component or gas is directly proportional to its mole fraction in the solution.

Teacher's Note:
1. Raoult’s law is the special case of Henry’s law in which K_H = p°.

Question 7. What type of liquids form ideal solutions?
Answer:
Liquids that have similar structures and polarities form ideal solutions.

Teacher's Note:
1. Example: benzene and toluene, or n-hexane and n-heptane.

Question 8. Under what condition do non-ideal solutions show negative deviations?
Answer:
When the new forces of interaction between the components are stronger than those in the pure components, then non-ideal solutions show negative deviations.

Teacher's Note:
1. Stronger A–B forces mean fewer molecules escape, so the observed vapour pressure is lower than Raoult’s law predicts.

Question 9. What are maximum boiling azeotropes? Give one example.
Answer:
Maximum boiling azeotropes are those which boil at higher temperature than boiling point of each component in pure state, e.g., 68% nitric acid and 32% water by mass.

Teacher's Note:
1. Maximum boiling azeotropes come from negative deviation; minimum boiling ones (like 95% ethanol–water) come from positive deviation.

Question 10. Why is the vapour pressure of a solution of glucose in water lower than that of water? [NCERT Exemplar]
Answer:
This is due to decrease in the escaping tendency of the water molecules from the surface of solution as some of the surface area is occupied by non-volatile solute, glucose particles.

Teacher's Note:
1. Glucose is non-volatile, so only water molecules contribute to the vapour pressure of the solution.

Question 11. Define molal depression constant or cryoscopic constant.
Answer:
It is the depression in freezing point when 1 mole of non-volatile solute is dissolved in 1000 g of solvent.

Teacher's Note:
1. Its unit is K kg mol⁻¹, and it depends only on the solvent, not on the solute.

Question 12. What do you understand by the term that K_f for water is 1.86 K kg mol⁻¹?
Answer:
It means that the freezing point of water is lowered by 1.86 K when 1 mole of non-volatile solute is dissolved in 1 kg of water.

Teacher's Note:
1. This applies to a non-electrolyte; for an electrolyte multiply by the van’t Hoff factor i.

Question 13. What is an antifreeze?
Answer:
A substance such as ethylene glycol which is added to water to lower its freezing point is called an antifreeze. It is named so as it delays freezing.

Teacher's Note:
1. Antifreeze is an everyday application of depression in freezing point.

Question 14. What is de-icing agent? How does it work?
Answer:
Common salt is called de-icing agent as it lowers the freezing point of water to such an extent that it does not freeze to form ice. Thus, it is used to clear snow from roads.

Teacher's Note:
1. NaCl gives two ions per formula unit, so it depresses the freezing point about twice as much as a non-electrolyte of the same molality.

Short Answer Questions (2 Marks)

Question 15. Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions.
Answer:
A substance (solute) dissolves in a solvent if the intermolecular interactions are similar in both the components; for example, polar solutes dissolve in polar solvents and non polar solutes in non polar solvents and thus we can say “like dissolves like”.

Teacher's Note:
1. Give an example of each type: NaCl or sugar in water (polar), naphthalene or anthracene in benzene (non-polar).

Question 16. State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid? [CBSE (F) 2016]
Answer:
It states that the partial pressure of a gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
p ∝ x or p = K_Hx, where K_H is the Henry’s constant.
Application of Henry’s law: To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.
Effect of temperature on solubility: As dissolution is an exothermic process, therefore, according to Le Chatelier’s principle solubility should decrease with rise in temperature.

Teacher's Note:
1. Answer all three parts separately – law, application and temperature effect each carry marks.

Question 17. Henry’s law constant (K_H) for the solution 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. [CBSE (F) 2013]
Answer:
Here, K_H = 4.27 × 10⁵ mm Hg, p = 760 mm Hg
According to Henry’s law, p = K_Hx_CH4
∴ \( x_{CH_4} = \frac{p}{K_H} = \frac{760\text{ mm Hg}}{4.27\times10^{5}\text{ mm Hg}} = 1.78\times10^{-3} \)
Mole fraction of methane in benzene, x_CH4 = 1.78 × 10⁻³.

Teacher's Note:
1. K_H and p must be in the same unit; the mole fraction itself has no unit.

Question 18. State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law? [CBSE Delhi 2014; 2020 (56/5/1)]
Answer:
It states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. According to Raoult’s law, for a volatile component A of the solution p_A ∝ x_A or p_A = p_A°x_A, where p_A° is the vapour pressure of pure component A.
If one of the component is so volatile that it exist as a gas then according to Henry’s law p = K_Hx, where K_H is the Henry law constant i.e., the partial vapour pressure of the volatile component (gas) is directly proportional to its mole fraction in the solution.
Thus the similarity between Raoult’s law and Henry’s law is that in both the laws, the partial vapour pressure of the volatile component or gas is directly proportional to its mole fraction in the solution.

Teacher's Note:
1. The only difference is the proportionality constant: p° in Raoult’s law and K_H in Henry’s law.

Question 19. State the following:
(i) Raoult’s law in its general form in reference to solutions.
(ii) Henry’s law about partial pressure of a gas in a mixture. [CBSE (AI) 2011]
Answer:
(i) Raoult’s law: It states that for any solution, the partial pressure of each volatile component in the solution is directly proportional to its mole fraction.
(ii) Henry’s law: It states that the partial pressure of a gas in vapour phase (p) is proportional to its mole fraction (x) in the solution.

Teacher's Note:
1. Write the mathematical forms too: p_A = p_A°x_A and p = K_Hx.

Question 20. Define an ideal solution and write one of its characteristics. [CBSE Delhi 2014]
Answer:
A solution which obeys Raoult’s law over the entire range of concentration is called ideal solution. The important characteristics of an ideal solution are
(i) The enthalpy of mixing of pure components to form the solution is zero i.e., Δ_mixH = 0
(ii) The volume of mixing is zero i.e., Δ_mixV = 0

Teacher's Note:
1. “Over the entire range of concentration” is the key phrase in the definition – do not leave it out.

Teacher's Note:
1. Strictly, only non-ideal solutions with large deviations form azeotropes; the ΔH and ΔV points are the safest differences to quote.

Teacher's Note:
1. Positive deviation: weaker A–B forces, heat absorbed, volume expands. Negative deviation: the exact opposite on every point.

Question 23. State Raoult’s law for a solution containing non-volatile solute. What type of deviation from Raoult’s law is shown by a solution of chloroform and acetone and why? [CBSE (F) 2017]
Answer:
It states that the relative lowering of vapour pressure is equal to mole fraction of solute when solvent alone is volatile and is expressed as
\( \frac{p_A^\circ - p_{Total}}{p_A^\circ} = x_B \)
A solution of chloroform and acetone shows negative deviation from Raoult’s law. This is because chloroform molecule is able to form H–bond with acetone molecule as shown below.
(CH₃)₂C=O ······ H–CCl₃
It decreases the escaping tendency of molecules of each component from the surface of solution and consequently the vapour pressure decreases resulting in negative deviation from Raoult’s law.

Teacher's Note:
1. The three Cl atoms make the C–H hydrogen of chloroform slightly acidic, which is why it can H-bond with the oxygen of acetone.

Question 24. (i) Why is an increase in temperature observed on mixing chloroform and acetone? [CBSE 2019 (56/2/3)]
(ii) Why does sodium chloride solution freeze at a lower temperature than water? [CBSE (F) 2013]
Answer:
(i) The bonds between chloroform molecules and molecules of acetone are dipole-dipole interactions but on mixing, the chloroform and acetone molecules, they start forming hydrogen bonds which are stronger bonds resulting in the release of energy. This gives rise to an increase in temperature.
(ii) When a non-volatile solute is dissolved in a solvent, the vapour pressure decreases. As a result, the solvent freezes at a lower temperature.

Teacher's Note:
1. Heat released on mixing means Δ_mixH < 0, which is a sign of negative deviation.

Question 25. Define azeotropes. What type of azeotrope is formed by negative deviation from Raoult’s law? Give an example. [CBSE Delhi 2015]
Answer:
Azeotropes are binary liquid mixtures having the same composition in liquid and vapour phase and boil at a constant temperature.
Maximum boiling azeotrope is formed by negative deviation from Raoult’s law. A mixture of 68% nitric acid and 32% water by mass is an example of maximum boiling azeotrope.

Teacher's Note:
1. Because liquid and vapour have the same composition, the components of an azeotrope cannot be separated by fractional distillation.

Question 26. What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is formed by mixing ethanol and acetone? [CBSE (F) 2013]
Answer:
A mixture of ethanol and acetone shows positive deviation and the azeotrope formed by this mixture is minimum boiling azeotrope.

Teacher's Note:
1. Give the reason: acetone molecules get between ethanol molecules and break some of their hydrogen bonds, weakening the interactions.

Question 27. (i) Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of K_H (Henry’s constant) and why?
(ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes? [CBSE Central 2016]
Answer:
(i) According to Henry’s law, p = K_Hx, i.e., higher the value of K_H lower is the solubility of the gas in the liquid. Therefore, Gas B will have higher value of K_H than gas A.
(ii) Negative deviation from Raoult’s law.

Teacher's Note:
1. Remember the inverse relation: more soluble gas → smaller K_H.

Short Answer Questions (3 Marks)

Question 28. At 25°C the saturated vapour pressure of water is 3.165 kPa (23.75 mm Hg). Find the saturated vapour pressure of a 5% aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea = 60.05 g mol⁻¹) [CBSE (F) 2012]
Answer:
W_B = 5 g, W_A = 95 g, M_B = 60.05 g mol⁻¹, M_A = 18 g mol⁻¹, p°_A = 3.165 kPa
Substituting the values in the expression
\( \frac{p_A^\circ - p}{p_A^\circ} = \frac{W_B\times M_A}{M_B\times W_A} \), we get
\( \frac{3.165\text{ kPa} - p}{3.165\text{ kPa}} = \frac{5\text{ g}\times18\text{ g mol}^{-1}}{60.05\text{ g mol}^{-1}\times95\text{ g}} \) = 0.0158
p = 3.165 kPa − 0.0158 × 3.165 kPa
p = 3.115 kPa

Teacher's Note:
1. A 5% (w/w) solution means 5 g solute in 100 g solution, i.e. 95 g of water – not 100 g of water.

Question 29. Calculate the boiling point of solution when 2 g of Na₂SO₄ (M = 142 g mol⁻¹) was dissolved in 50 g of water, assuming Na₂SO₄ undergoes complete ionisation. (K_b for water = 0.52 K kg mol⁻¹) [CBSE North 2016]
Answer:
i = \( \frac{3}{1} \) = 3   (Na₂SO₄ → 2Na⁺ + SO₄²⁻)
Given, W_B = 2 g, W_A = 50 g, M_B = 142 g mol⁻¹, K_b = 0.52 K kg mol⁻¹
Substituting the values in the expression, \( \Delta T_b = \frac{i\times K_b\times W_B\times1000}{M_B\times W_A} \)
\( \Delta T_b = \frac{3\times0.52\times2\times1000}{142\times50} \) = 0.439 K
So, boiling point of solution, T_b = T_b⁰ + ΔT_b
= 373.15 K + 0.439 K = 373.589 K

Teacher's Note:
1. Count ions carefully: Na₂SO₄ gives three ions (2Na⁺ + SO₄²⁻), so i = 3, not 2.

Question 30. A solution of glucose (Molar mass = 180 g mol⁻¹) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water K_f and K_b are 1.86 K kg mol⁻¹ and 0.512 K kg mol⁻¹ respectively. [CBSE (F) 2017]
Answer:
ΔT_b = 100.20°C − 100°C = 0.20°C or 0.20 K
ΔT_b = K_b . m   ⇒   m = ΔT_b/K_b
m = \( \frac{0.20\text{ K}}{0.512\text{ K kg mol}^{-1}} \) or m = 0.39 mol kg⁻¹
ΔT_f = K_f m
ΔT_f = 1.86 K kg mol⁻¹ × 0.39 mol kg⁻¹ = 0.725 K
T_f = T_f⁰ − ΔT_f
= 273.15 K − 0.725 K = 272.425 K

Teacher's Note:
1. Shortcut: ΔT_f = ΔT_b × K_f/K_b, because the molality is the same for both; the molar mass is not needed.

Question 31. Calculate the freezing point of a solution when 3 g of CaCl₂ (M = 111 g mol⁻¹) was dissolved in 100 g of water, assuming CaCl₂ undergoes complete ionisation. (K_f for water = 1.86 K kg mol⁻¹) [CBSE East 2016]
Answer:
Here, W_B = 3 g, i = \( \frac{3}{1} \) = 3 (CaCl₂ → Ca²⁺ + 2Cl⁻), W_A = 100 g, K_f = 1.86 K kg mol⁻¹, M_B = 111 g mol⁻¹
Substituting these values in the expression,
\( \Delta T_f = \frac{i\times K_f\times W_B\times1000}{M_B\times W_A} \), we get
\( \Delta T_f = \frac{3\times1.86\times3\times1000}{111\times100} \) = 1.508 K
Freezing point of solution,
T_f = T_f° − ΔT_f
= 273.15 K − 1.508 K = 271.642 K

Teacher's Note:
1. Freezing point of the solution is always lower than that of pure water – subtract ΔT_f, never add it.

Question 32. 3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van’t Hoff factor and predict the nature of solute (associated or dissociated). (Given: Molar mass of benzoic acid = 122 g mol⁻¹, K_f for benzene = 4.9 K kg mol⁻¹) [CBSE Delhi 2015]
Answer:
The given quantities are
W_B = 3.9 g, W_A = 49 g, ΔT_f = 1.62 K, M_B = 122 g mol⁻¹ and K_f = 4.9 K kg mol⁻¹
Substituting these values in the equation,
\( M_B = \frac{K_f\times W_B\times1000}{\Delta T_f\times W_A} \), we get
\( M_B = \frac{4.9\times3.9\times1000}{1.62\times49} \)
M_B = 240.74 g mol⁻¹
Thus, observed molecular mass of benzoic acid in benzene = 240.74 g mol⁻¹
Normal molecular mass of benzoic acid = 122 g mol⁻¹
i = \( \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}} \)
i = \( \frac{122\text{ g mol}^{-1}}{240.74\text{ g mol}^{-1}} \) = 0.507
As i < 1, therefore, benzoic acid will undergo association in benzene.

Teacher's Note:
1. Benzoic acid forms hydrogen-bonded dimers in benzene, which is why the observed molar mass is nearly double (about 244 for a perfect dimer).

Question 33. Calculate the mass of NaCl (molar mass = 58.5 g mol⁻¹) to be dissolved in 37.2 g of water to lower the freezing point by 2°C, assuming that NaCl undergoes complete dissociation. (K_f for water = 1.86 K kg mol⁻¹) [CBSE (F) 2015]
Answer:
NaCl → Na⁺ + Cl⁻, i = \( \frac{2}{1} \) = 2
Substituting K_f = 1.86 K kg mol⁻¹, W_A = 37.2 g, M_B = 58.5 g mol⁻¹, i = 2, ΔT_f = 2°C or 2 K in the equation,
\( W_B = \frac{\Delta T_f\times M_B\times W_A}{i\times K_f\times1000} \), we get
\( W_B = \frac{2\times58.5\times37.2}{2\times1.86\times1000} \) = 1.17 g

Teacher's Note:
1. A change of 2°C equals a change of 2 K – temperature differences need no conversion.

Question 34. At 300 K, 30 g of glucose, C₆H₁₂O₆ present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution. [CBSE 2019 (56/4/2)]
Answer:
\( \pi = \frac{W_B\times R\times T}{M_B\times V} = CRT \)
\( 4.98 = \frac{30\times R\times300}{180\times1} \) = 50R   ...(i)
1.52 = C × R × 300 = 300CR   ...(ii)
Dividing equation (ii) by (i), we get
\( \frac{300CR}{50R} = \frac{1.52}{4.98} \)   ⇒   C = 0.051 M

Teacher's Note:
1. Dividing the two equations cancels R and T, so you never need the value of R here.

Question 35. A solution containing 1.9 g per 100 mL of KCl (M = 74.5 g mol⁻¹) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol⁻¹). Calculate the degree of dissociation of KCl solution. Assume that both the solutions have same temperature. [CBSE 2019 (56/2/1)]
Answer:
Isotonic solutions are those solutions which have the same osmotic pressure. If the osmotic pressures are equal at the same temperature, concentrations must also be equal. So,
For non-electrolyte Urea; π = CRT
For electrolyte KCl; π = i CRT
Thus, C_ureaRT = iC_KClRT
C_urea = iC_KCl
C_urea = \( \frac{3\times1000}{60\times100} \) = 0.5 M
C_KCl = \( \frac{1.9\times1000}{74.5\times100} \) = 0.25 M
So, 0.5 = i × 0.25
i = 2.0
Now, KCl → K⁺ + Cl⁻
Initial: 0.25, 0, 0; at equilibrium: 0.25 − α, α, α
Total moles after dissociation = 0.25 − α + α + α
i = \( \frac{\text{Total number of moles after dissociation}}{\text{Total number of moles before dissociation}} \)
i = \( \frac{0.25+\alpha}{0.25} \)   ⇒   2.0 = \( \frac{0.25+\alpha}{0.25} \)
α = 0.25 or α = 25%

Teacher's Note:
1. Here α is written in mol L⁻¹ of KCl dissociated; the fraction dissociated is 0.25/0.25 = 1. Using i = 1 + α directly, i = 2 gives α = 1, i.e. complete dissociation.

Long Answer Questions (5 Marks)

Question 36. (i) Define the following terms:
(a) Molarity
(b) Molal elevation constant (K_b)
(ii) A solution containing 15 g urea (molar mass = 60 g mol⁻¹) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol⁻¹) in water. Calculate the mass of glucose present in one litre of its solution. [CBSE (AI) 2014]
Answer:
(i) (a) Molarity may be defined as number of moles of solute dissolved in one litre of solution.
Molarity = \( \frac{\text{Moles of solute}}{\text{Volume of solution (in litre)}} \)
(b) Molal elevation constant may be defined as the elevation in boiling point when one mole of solute is dissolved in 1000 grams of the solvent.
(ii) Osmotic pressure, \( \pi = \frac{W_B\times R\times T}{M_B\times V} \)
Osmotic pressure of urea solution = \( \frac{15\times R\times T}{60\times1} \)
Osmotic pressure of glucose solution = \( \frac{W_B\times R\times T}{180\times1} \)
As, Osmotic pressure of urea solution = Osmotic pressure of glucose solution
Therefore, \( \frac{15\times R\times T}{60\times1} = \frac{W_B\times R\times T}{180\times1} \)
Mass of glucose, \( W_B = \frac{15\times180}{60} \) = 45 g

Teacher's Note:
1. Isotonic solutions of non-electrolytes have equal molar concentrations: 15/60 = 0.25 M, so glucose needed = 0.25 × 180 = 45 g.

Question 37. (i) Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases.
(ii) Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the number of moles of CO₂ in 500 ml of soda water when packed under 2.53 × 10⁵ Pa at the same temperature. [CBSE (AI) 2017C]
Answer:
(i) On addition of glucose, a non-volatile solute to water (a volatile solvent) the vapour pressure of water decreases. In order to make the solution boil, its vapour pressure must be increased by raising the temperature above the boiling point of pure water. In other words, there is an increase in the boiling point of water.
(ii) According to Henry’s law, p_CO2 = K_H × x_CO2 or \( x_{CO_2} = \frac{p_{CO_2}}{K_H} \)
∴ \( x_{CO_2} = \frac{2.53\times10^5\text{ Pa}}{1.67\times10^8\text{ Pa}} \) = 1.51 × 10⁻³
Mass of water = Density of water × volume of water
= 1 g mL⁻¹ × 500 mL = 500 g
Moles of water, n = \( \frac{\text{Mass of water}}{\text{Molar mass}} \) = \( \frac{500\text{ g}}{18\text{ g mol}^{-1}} \) = 27.78 mol
\( x_{CO_2} = \frac{n_{CO_2}}{n_{H_2O}+n_{CO_2}} \approx \frac{n_{CO_2}}{n_{H_2O}} \) or n_CO2 = x_CO2 × n_H2O
n_CO2 = 1.51 × 10⁻³ × 27.78 mol
= 0.042 mol

Teacher's Note:
1. Since CO₂ is only slightly soluble, n_CO2 is neglected beside n_H2O in the denominator; keep both pressures in the same unit (Pa).

Question 38. (i) Define the following terms:
(a) Azeotrope
(b) Osmotic pressure
(c) Colligative properties
(ii) Calculate the molarity of 9.8% (w/w) solution of H₂SO₄ if the density of the solution is 1.02 g mL⁻¹. (Molar mass of H₂SO₄ = 98 g mol⁻¹) [CBSE (F) 2014]
Answer:
(i) (a) The binary mixtures of liquids having same composition in liquid and vapour phase and boil at a constant temperature are called azeotropes.
(b) The excess of pressure which must be applied to the solution side to prevent the passage of solvent into it through a semipermeable membrane is called osmotic pressure.
(c) The properties of solutions which depend only on the number of solute particles in the solution but independent of their nature are called colligative properties.
(ii) Let the mass of solution = 100 g
∴ Mass of H₂SO₄ = 9.8 g
Number of mole of H₂SO₄ = \( \frac{\text{Mass of H}_2\text{SO}_4}{\text{Molar Mass}} \) = \( \frac{9.8\text{ g}}{98\text{ g mol}^{-1}} \) = 0.1 mol
Volume of solution = \( \frac{\text{Mass of solution}}{\text{Density of solution}} \) = \( \frac{100\text{ g}}{1.02\text{ g mL}^{-1}} = \frac{100}{1.02} \) mL
= \( \frac{100}{1.02}\text{ mL}\times\frac{1}{1000\text{ mL L}^{-1}} = \frac{1}{10.2} \) L
Molarity = \( \frac{\text{Moles of solute}}{\text{Volume of solution (in L)}} \)
= \( \frac{0.1\text{ mol}}{\frac{1}{10.2}\text{ L}} \) = 1.02 mol L⁻¹ or 1.02 M

Teacher's Note:
1. For w/w % with density, take 100 g of solution and convert its mass to volume using the density – do not use 100 mL directly.

Question 39. Discuss biological and industrial importance of osmosis.
Answer:
The process of osmosis is of great biological and industrial importance as is evident from the following examples:
(i) Movement of water from soil into plant roots and subsequently into upper portion of the plant occurs partly due to osmosis.
(ii) Preservation of meat against bacterial action by adding salt.
(iii) Preservation of fruits against bacterial action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
(iv) Reverse osmosis is used for desalination of water.

Teacher's Note:
1. For a 5-mark answer, add one line of explanation to each example (for salting and sugaring, the bacterium loses water to the hypertonic medium).

Question 40. (i) When 2.56 g of sulphur was dissolved in 100 g of CS₂, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_x). [K_f for CS₂ = 3.83 K kg mol⁻¹, Atomic mass of Sulphur = 32 g mol⁻¹]
(ii) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(a) 1.2% sodium chloride solution?
(b) 0.4% sodium chloride solution? [CBSE Delhi 2016]
Answer:
(i) Here, W_B = 2.56 g, W_A = 100 g, ΔT_f = 0.383 K, K_f = 3.83 K kg mol⁻¹
Substituting these values in the expression, we get
\( M_B = \frac{K_f\times W_B\times1000}{\Delta T_f\times W_A} \)
\( M_B = \frac{3.83\times2.56\times1000}{100\times0.383} \) = 256 g mol⁻¹
Now, molecular mass of S_x = x × 32 = 256
x = \( \frac{256}{32} \) = 8
Therefore, formula of sulphur = S₈
(ii) (a) Water will flow out of the cell and they would shrink.
(b) Water will flow into the cell and they would swell.

Teacher's Note:
1. 1.2% NaCl is hypertonic and 0.4% NaCl is hypotonic compared with the 0.9% (isotonic) solution; water always moves towards the more concentrated side.