Question

Acetic acid dimerizes when dissolved in benzene. As a result boiling point of the solution rises by \( 0.36^\circ \text{C} \), when 100 g of benzene is mixed with "X" g of acetic acid. In this solution, if experimentally measured molecular weight of acetic acid is 117.8 and molar elevation constant of benzene is 2.57 K kg mol\(^{-1} \), what is the weight % and degree of dissociation (in %) of acetic acid in benzene?

• A. 1.62 and 98.3
• B. 0.81 and 98.3
• C. 0.5 and 86
• D. 1 and 98.3
• E. 1.4 and 99

Hint:

For association (dimerization), \( i < 1 \). Always use \( i = \frac{\text{normal molar mass}}{\text{observed molar mass}} \).

Answer 1

The Correct Option is A

Concept: Acetic acid dimerizes in benzene, so van’t Hoff factor: \[ i = 1 - \frac{\alpha}{2} \] Also, \[ \Delta T_b = i K_b m \quad \text{and} \quad i = \frac{\text{Normal molar mass}}{\text{Observed molar mass}} \]

Step 1: {Calculate van’t Hoff factor.}
Normal molar mass of acetic acid = 60 \[ i = \frac{60}{117.8} \approx 0.509 \]

Step 2: {Find degree of dimerization.}
\[ i = 1 - \frac{\alpha}{2} \] \[ 0.509 = 1 - \frac{\alpha}{2} \] \[ \frac{\alpha}{2} = 0.491 \] \[ \alpha = 0.982 \approx 98.2\% \]

Step 3: {Use boiling point elevation to find molality.}
\[ \Delta T_b = i K_b m \] \[ 0.36 = (0.509)(2.57)m \] \[ m = \frac{0.36}{0.509 \times 2.57} \approx 0.275 \, \text{mol/kg} \]

Step 4: {Find mass of acetic acid.}
Molality: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] Solvent = 100 g = 0.1 kg \[ \text{moles} = 0.275 \times 0.1 = 0.0275 \] \[ \text{mass} = 0.0275 \times 60 = 1.65 \, \text{g} \]

Step 5: {Calculate weight %.}
\[ \text{Weight \%} = \frac{1.65}{100 + 1.65} \times 100 \] \[ \approx 1.62\% \]

Step 6: {Final answers.}
\[ \boxed{\text{Weight \%} \approx 1.62\%} \] \[ \boxed{\text{Degree of dimerization} \approx 98.2\%} \]