Equimolal solutions of KCl and compound X in water shows depressions in freezing point in the ratio of 4: 1 . Assuming KCl to be completely ionised, the compound X in solution must

Question

Equimolal solutions of KCl and compound X in water shows depressions in freezing point in the ratio of 4: 1 . Assuming KCl to be completely ionised, the compound X in solution must (A) dissociate to the extent of 50 % (B) hydrolyse to the extent of 80 % (C) dimerise to the extent of 50 % (D) trimérise to the extent of 75 %

Tardigrade Admin · Accepted Answer

(Δ Tf( KCl )/Δ Tf(X))=(4/1)=(i( KCl )/i(X))=(2/i(X)) ⇒ i(X)=0.5 For association of 3 molecules, i(X)=1-(1-(1/n)) α=1-(1-(1/3)) α=0.5 ⇒ α=0.75

Q. Equimolal solutions of $K Cl$ and compound $X$ in water shows depressions in freezing point in the ratio of $4 : 1$ . Assuming $K Cl$ to be completely ionised, the compound $X$ in solution must

dissociate to the extent of $50%$

hydrolyse to the extent of $80%$

dimerise to the extent of $50%$

trimérise to the extent of $75%$

$\frac{Δ T _{f} ( K Cl )}{Δ T _{f} ( X )} = \frac{4}{1} = \frac{i ( K Cl )}{i ( X )} = \frac{2}{i ( X )}$

$\Rightarrow i (X) = 0.5$

For association of 3 molecules,

$i (X) = 1 - (1 - \frac{1}{n}) α = 1 - (1 - \frac{1}{3}) α = 0.5$

$\Rightarrow α = 0.75$

Q. Equimolal solutions of KCl and compound X in water shows depressions in freezing point in the ratio of 4:1 . Assuming KCl to be completely ionised, the compound X in solution must

dissociate to the extent of 50%

hydrolyse to the extent of 80%

dimerise to the extent of 50%

trimérise to the extent of 75%