LEVEL 1 WORKSHEET CHAPTER SOLUTION

 Class 12 Chemistry Chapter – Solutions, based on NCERT and CBSE Board Exam patterns, along with answers:


1. Define mole fraction. How is it calculated?

Answer:
Mole fraction of a component in a solution is the ratio of number of moles of that component to the total number of moles of all components present in the solution.

Mole fraction of A, xA=nAnA+nB\text{Mole fraction of A, } x_A = \frac{n_A}{n_A + n_B}

2. State Henry’s Law. Write its two applications.

Answer:
Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the liquid.

C=kHPC = k_H \cdot P

Where
= concentration of gas,
= partial pressure, kHk_H = Henry's constant.

Applications:

  1. In carbonated beverages (soda), CO₂ is dissolved under high pressure.

  2. Scuba divers experience bends due to release of N₂ gas dissolved under high pressure.


3. What are ideal and non-ideal solutions? Give one example of each.

Answer:

  • Ideal Solution: A solution that obeys Raoult’s law at all concentrations.
    Example: Benzene + Toluene

  • Non-Ideal Solution: A solution that does not obey Raoult’s law.
    Example: Ethanol + Acetone


4. What is Raoult’s Law? Write its mathematical expression for a binary solution.

Answer:
Raoult’s Law states that the partial vapor pressure of a component in a solution is directly proportional to its mole fraction.

For component A and B:

PA=xAPA0andPB=xBPB0P_A = x_A P_A^0 \quad \text{and} \quad P_B = x_B P_B^0

Total pressure:

Ptotal=PA+PB=xAPA0+xBPB0P_{total} = P_A + P_B = x_A P_A^0 + x_B P_B^0

5. What is an azeotrope? Name two types with one example each.

Answer:
Azeotropes are binary mixtures that boil at constant temperature and cannot be separated by fractional distillation.

Types:

  1. Minimum boiling azeotrope – Ethanol + Water

  2. Maximum boiling azeotrope – HCl + Water


6. Define colligative properties. Name four such properties.

Answer:
Colligative properties are properties that depend on the number of solute particles, not their nature.

Examples:

  1. Relative lowering of vapor pressure

  2. Elevation of boiling point

  3. Depression of freezing point

  4. Osmotic pressure


7. Derive the formula for depression in freezing point.

Answer:

ΔTf=Kfm\Delta T_f = K_f \cdot m

Where:

  • ΔTf\Delta T_f = depression in freezing point

  • KfK_f = cryoscopic constant

  • mm = molality of solution


8. Define van’t Hoff factor (i). What is its significance?

Answer:
Van’t Hoff factor (i) is the ratio of the actual number of particles in solution after dissociation/association to the number of formula units dissolved.

i=Observed colligative propertyCalculated colligative property (assuming no dissociation/association)i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property (assuming no dissociation/association)}}

Significance: It accounts for ionisation/dissociation (i > 1) and association (i < 1).


9. Calculate the osmotic pressure of a solution prepared by dissolving 0.5 mol of glucose in 1 L of water at 27°C. (R = 0.0821 L·atm/mol·K)

Answer:

π=CRT=(0.5)(0.0821)(300)=12.315atm\pi = C R T = (0.5)(0.0821)(300) = 12.315 \, \text{atm}

10. A solution of urea (mol. mass = 60 g/mol) is prepared by dissolving 18 g of urea in 100 g of water. Calculate the lowering in vapor pressure if vapor pressure of pure water at 298 K is 23.8 mmHg.

Answer:

Moles of urea=1860=0.3

\text{Moles of urea} = \frac{18}{60} = 0.3





















Moles of water=100185.56
\text{Moles of water} = \frac{100}{18} \approx 5.56
xurea=0.30.3+5.560.051x_{\text{urea}} = \frac{0.3}{0.3 + 5.56} \approx 0.051 Lowering in vapor pressure=P0xurea=23.80.0511.21mmHg\text{Lowering in vapor pressure} = P^0 \cdot x_{\text{urea}} = 23.8 \cdot 0.051 \approx 1.21 \, \text{mmHg}

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