JEE Main · 2017 · 02 Apr 2017 · Offline · Solutions

The freezing point of benzene decreases by

0.4 5^{\circ} C

when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be :

(K_{f}

for benzene

= 5.12 K kg mol^{- 1})

(A) $64.6%$
(B) $80.4%$
(C) $74.6%$
(D) $94.6%$ ✓ Correct

Answer: (D)

Solution

The depression is

Δ T_{f} = 0.45

K. Without association the molality is

\frac{0.2/60}{0.020} = 0.167

mol kg

^{- 1}

, which would depress the freezing point by

K_{f} m = 5.12 \times 0.167 = 0.853

K. The observed

0.45

K is smaller, giving a van't Hoff factor

i = \frac{0.45}{0.853} = 0.527

. For dimerisation

i = 1 - \frac{α}{2}

, so

\frac{α}{2} = 0.473

and

α \approx 0.946

— about

94.6%

of the acid pairs into dimers.

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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