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

Determine the degree of association (polymerization) for the reaction in aqueous solution $6 HCHO \rightleftharpoons C_6H_{12}O_6$. If the observed (mean) molar mass of the mixture is $150 \ g/mol$. (Given: Molar mass of $HCHO = 30 \ g/mol$)

$1.2 \ g$ of $NaCl$ solution is isotonic with $7.2 \ g$ of glucose solution. Calculate the van’t Hoff factor $(i)$ of the $NaCl$ solution.

Which of the following expressions correctly defines the Van't Hoff factor $i$?

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

When $20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ is dissolved in $50 \ g$ of benzene $(K_{f} = 1.72 \ K \ kg \ mol^{-1})$,a freezing point depression of $2 \ K$ is observed. The van't Hoff factor $(i)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module