JEE MAIN - Chemistry (2017 (Offline) - No. 22)
The freezing point of benzene decreases by 0.450C when 0.2 g 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 = 5.12 K kg mol–1)
80.4 %
74.6 %
94.6 %
64.6 %
Explanation
$$\Delta $$Tf = i $$ \times $$ Kf $$ \times $$ m
$$ \Rightarrow $$ 0.45 = i $$ \times $$ 5.12 $$ \times $$ $${{0.2 \times 1000} \over {60 \times 20}}$$
$$ \Rightarrow $$ i = 0.527
2CH3COOH ⇌ (CH3COOH)2
1 - $$\alpha $$ $${\alpha \over 2}$$
i = 1 - $$\alpha $$ + $${\alpha \over 2}$$
$$ \Rightarrow $$ $$\alpha $$ = 0.946
$$ \therefore $$ % dissociation is 94.6%.
$$ \Rightarrow $$ 0.45 = i $$ \times $$ 5.12 $$ \times $$ $${{0.2 \times 1000} \over {60 \times 20}}$$
$$ \Rightarrow $$ i = 0.527
2CH3COOH ⇌ (CH3COOH)2
1 - $$\alpha $$ $${\alpha \over 2}$$
i = 1 - $$\alpha $$ + $${\alpha \over 2}$$
$$ \Rightarrow $$ $$\alpha $$ = 0.946
$$ \therefore $$ % dissociation is 94.6%.
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