The freezing point of benzene decreases by 0.45° 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)

Question

The freezing point of benzene decreases by 0.45° 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) 94.6 % (B) 64.6 % (C) 80.4 % (D) 74.6 %

Tardigrade Admin · Accepted Answer

Δ Tf=i× m× kf 0.45=i× (0 . 2 × 1000/60 × 20)× 5.12 i=0.527=1-α +(α /n) for association 0.527=1-α +(α /2)=1-(α /2) (α /2)=1-0.527=0.473α =0.946 %α =94.6

Q. 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 $= 5.12 K k g m o l^{- 1})$

$94.6%$

$64.6%$

$80.4%$

$74.6%$

$Δ T_{f} = i \times m \times k_{f}$
$0.45 = i \times \frac{0.2 \times 1000}{60 \times 20} \times 5.12$
$i = 0.527 = 1 - α + \frac{α}{n}$ for association
$0.527 = 1 - α + \frac{α}{2} = 1 - \frac{α}{2}$
$\frac{α}{2} = 1 - 0.527 = 0.473 α = 0.946% α = 94.6$

Q. The freezing point of benzene decreases by 0.45∘C when 0.2g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: Kf​ for benzene =5.12Kkgmol−1)

94.6%

64.6%

80.4%

74.6%