Question

Benzoic acid undergoes dimerization in benzene. \( x \) g of benzoic acid (molar mass 122 g mol\(^{-1}\)) is dissolved in 49 g of benzene. The depression in freezing point is 1.12 K. If degree of association of acid is 88%, what is the value of \( x \)? (K\(_f\) for benzene = 4.9 K kg mol\(^{-1}\))

• A. 2.44
• B. 1.22
• C. 3.66
• D. 11.36

Hint:

Remember to account for the degree of association when calculating molality. The equation used for depression in freezing point helps to calculate the exact value of the solute.

Answer 1

The Correct Option is D

Step 1: The depression in freezing point (\( \Delta T_f \)) is related to the molality (\( m \)) of the solution by the formula: \[ \Delta T_f = K_f \times m \] where: - \( \Delta T_f = 1.12 \, K \) is the depression in freezing point, - \( K_f = 4.9 \, K \, \text{kg mol}^{-1} \) is the cryoscopic constant for benzene, - \( m \) is the molality of the solution. The molality \( m \) is defined as: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] In this case, we are given: - Molar mass of benzoic acid = 122 g/mol, - Mass of solvent (benzene) = 49 g = 0.049 kg, - Degree of association \( \alpha = 88% = 0.88 \). For benzoic acid, it undergoes dimerization, meaning 2 moles of benzoic acid combine to form a dimer. So, the number of moles of benzoic acid is reduced by the factor \( 1 - \alpha \).
Step 2: We can now write the molality \( m \) as: \[ m = \frac{n_{\text{solute}}}{m_{\text{solvent}}} = \frac{\frac{x}{122} \times (1 - \alpha)}{0.049} \] Substitute the known values: \[ m = \frac{\frac{x}{122} \times (1 - 0.88)}{0.049} = \frac{\frac{x}{122} \times 0.12}{0.049} \] Simplifying further: \[ m = \frac{x \times 0.12}{122 \times 0.049} = \frac{0.12x}{5.978} \]
Step 3: Now we use the depression in freezing point equation: \[ \Delta T_f = K_f \times m \] Substitute the known values: \[ 1.12 = 4.9 \times \frac{0.12x}{5.978} \] Solving for \( x \): \[ 1.12 = \frac{0.588x}{5.978} \] \[ x = \frac{1.12 \times 5.978}{0.588} \] \[ x = \frac{6.686}{0.588} \] \[ x = 11.36 \, \text{g} \] So, the value of \( x \) is 11.36g