Solute A associates in water. When 0.7 g of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.First,calculate the molality $(m)$: $m = \frac{0.7 \ g}{93 \ g \ m

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.First,calculate the molality $(m)$: $m = \frac{0.7 \ g}{93 \ g \ mol^{-1}} 	imes \frac{1000 \ g \ kg^{-1}}{42.0 \ g} = 0.1792 \ mol \ kg^{-1}$.Using $\Delta T_f = 0.2 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$,we find the van't Hoff factor $(i)$:$0.2 = i \cdot 1.86 \cdot 0.1792 \implies i = \frac{0.2}{1.86 \cdot 0.1792} \approx 0.60$.For the association reaction $2A ightleftharpoons A_2$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.Equating $1 - \frac{\alpha}{2} = 0.60$,we get $\frac{\alpha}{2} = 0.40$,so $\alpha = 0.80$.The percentage association is $80 \%$.

Similar Questions

If the degree of association is $70 \%$ for the reaction $2 A \rightleftharpoons (A)_2$,the van't-Hoff factor for the solute $A$ is

Dimerisation of solute molecules in low dielectric constant solvent is due to

The experimental depression in freezing point of a dilute solution is $0.025 \ K$. If the van't Hoff factor $(i)$ is $2.0$,the calculated depression in freezing point (in $K$) is

Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module