Question

Molecules of benzoic acid (C$_{6}H_{5}$COOH) dimerise in benzene. $'w' \,g$ of the acid dissolved in $30\, g$ of benzene shows a depression in freezing point equal to $2K$. If the percentage association of the acid to form dimer in the solution is $80$, then $w$ is : (Given that $K_{f} = 5 K\, kg\, mol^{-1} $, Molar mass of benzoic acid $= 122 \,g\, mol^{-1})$

• A. 1.8 g
• B. 2.4 g
• C. 1.0 g
• D. 1.5 g

Answer 1

The Correct Option is B

The problem involves determining the mass \(w \,g\) of benzoic acid dissolved in benzene that results in a certain depression of freezing point, considering the dimerization of benzoic acid molecules. Let's solve it step-by-step: 

Step 1: Understanding Dimerization and Depression in Freezing Point

• Benzoic acid (\(C_6H_5COOH\)) tends to dimerize in benzene, which affects the depression in freezing point.
• The degree of association (\(\alpha\)) is given as 80%.

Step 2: Calculate the Effective Molar Mass (M)

The formula for the effective molar mass in case of dimerization is: \(M_{\text{effective}} = \frac{M_0}{1 - \alpha/2}\), where \(M_0\) is the molar mass of benzoic acid, \(122 \, \text{g/mol}\).

• Given, \(\alpha = 0.8\), \(M_{\text{effective}} = \frac{122}{1 - 0.8/2} = \frac{122}{0.6}\)
• Calculating gives us \(M_{\text{effective}} = 203.33 \, \text{g/mol}\)

Step 3: Use the Depression of Freezing Point Formula

The formula for depression in freezing point is: \(\Delta T_f = \frac{K_f \times w}{M_{\text{effective}} \times W}\)

• Where \(\Delta T_f = 2 \, \text{K}\), \(K_f = 5 \, \text{K kg/mol}\), \(W = 30 \, \text{g} = 0.03 \, \text{kg}\)

Step 4: Rearrange and Solve for \(w\)

Substituting the values, we have: \(2 = \frac{5 \times w}{203.33 \times 0.03}\)

• Simplifying, \(w = \frac{2 \times 203.33 \times 0.03}{5}\)
• Calculating gives: \(w = 2.4 \, \text{g}\)

Conclusion: Correct Answer

The mass of benzoic acid that needs to be dissolved in benzene to produce the given depression in freezing point, considering 80% dimerization, is 2.4 g.