Worksheet class test Solutions

 

Here is a detailed answer to your questions from the chapter “Solutions”:

---

Conceptual & Numerical Questions

1. Define ideal and non-ideal solutions. Give one example of each.

• Ideal Solution: A solution that obeys Raoult’s Law over the entire range of concentrations and temperature. There is no enthalpy change (ΔH = 0) and no volume change (ΔV = 0) on mixing.

  • Example: Benzene + Toluene.

• Non-Ideal Solution: A solution that does not obey Raoult’s Law over the entire range. There is a change in enthalpy and/or volume during mixing.

  • Example: Acetone + Chloroform.

---

2. State Raoult’s law for a solution containing volatile components. How is it related to the lowering of vapour pressure?

• Raoult’s Law: The partial vapour pressure of each volatile component in a solution is directly proportional to its mole fraction.

  • For component A:
    
    Where is partial vapour pressure, is mole fraction, and is vapour pressure of pure A.

• Relation to Lowering of Vapour Pressure:

  \Delta P = P^0 - P = X_B P^0

Hence, adding a non-volatile solute lowers the vapour pressure of the solvent.

---

3. Differentiate between molarity and molality. Why is molality preferred over molarity in temperature-dependent calculations?

• Molality is preferred because it does not change with temperature, making it more reliable for thermodynamic calculations.

---

4. A solution is prepared by dissolving 10 g of a solute in 100 g of water. The boiling point elevation is found to be 0.52°C. Calculate the molar mass of the solute.
Given: Kb = 0.52 K kg mol⁻¹

Solution:

\Delta T_b = K_b \cdot m

0.52 = 0.52 \cdot \frac{n}{0.1} \Rightarrow \frac{n}{0.1} = 1 \Rightarrow n = 0.1 \text{ mol} ]

\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{10}{0.1} = \boxed{100 \text{ g/mol}}

---

5. What is van’t Hoff factor? How does it help in determining the degree of association or dissociation of a solute?

• Van’t Hoff factor (i): It is the ratio of the observed colligative property to the calculated colligative property assuming no association/dissociation.

  i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}

• Use:
  • For dissociation:
  • For association:

It helps determine the degree of dissociation (α) or association, indicating how many particles the solute actually contributes to the solution.

---

Special Questions on Degree of Association/Dissociation

6. Ethanoic acid in benzene shows van’t Hoff factor . Calculate the degree of association (α).

Ethanoic acid dimerizes:

2 \text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2

Let initial moles = 1.
Moles after association =

i = \frac{\text{Total moles after association}}{\text{Initial moles}} = 1 - \frac{\alpha}{2}

0.5 = 1 - \frac{\alpha}{2} \Rightarrow \frac{\alpha}{2} = 0.5 \Rightarrow \alpha = \boxed{1} ]

100% association

---

7. K₂SO₄ dissociates as:

\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}

Given:

• m = 0.1 mol/kg
• ΔTf = 0.558°C
• Kf = 1.86 K kg mol⁻¹

First, calculate i:

\Delta T_f = i \cdot K_f \cdot m
\Rightarrow i = \frac{0.558}{1.86 \cdot 0.1} = \frac{0.558}{0.186} = 3

Let degree of dissociation be α.
Initial: 1 mole
After dissociation:

• K⁺ = 2α,
• SO₄²⁻ = α,
• Undissociated = 1 – α

Total particles =

i = 1 + 2\alpha = 3 \Rightarrow \alpha = \boxed{1}

100% dissociation

---

Application-Based/Reasoning Questions

8. Why does NaCl elevate the boiling point more than urea at the same molal concentration?

• NaCl is an electrolyte, and it dissociates into two ions (Na⁺ and Cl⁻), doubling the number of solute particles.
• Urea is non-electrolyte; it does not dissociate, so the number of particles remains the same.

Since colligative properties depend on the number of solute particles, NaCl shows a greater effect (boiling point elevation, freezing point depression) due to van’t Hoff factor , while for urea, .

---

Let me know if you’d like this converted into a worksheet or formatted as a class handout!