A 0.2 , m aqueous solution of a weak acid (HX | vedclass

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

Question

(B) For a weak acid $HX$ undergoing dissociation: $HX \rightleftharpoons H^+ + X^-$. The van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degr

Vedclass · Accepted Answer

$ - 0.45 $

Vedclass · Answer

(B) For a weak acid $HX$ undergoing dissociation: $HX ightleftharpoons H^+ + X^-$. The van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation. Given $\alpha = 20 \% = 0.2$,so $i = 1 + 0.2 = 1.2$. The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Substituting the values: $\Delta T_f = 1.2 	imes 1.86 \, ^o C/m 	imes 0.2 \, m = 0.4464 \, ^o C$. The freezing point of the solution is $T_f = T_f^o - \Delta T_f = 0 \, ^o C - 0.4464 \, ^o C = - 0.4464 \, ^o C \approx - 0.45 \, ^o C$.

$A$ $0.2 \, m$ aqueous solution of a weak acid $(HX)$ is $20 \% $ ionised. The freezing point of this solution is .............. $^o C$ (Given ${K_f} = 1.86 \, ^o C/m$ for water)

Similar Questions

The van't Hoff factor $(i)$ for the reaction of $HNO_3$ with $H_2SO_4$ is:

At $25 \ ^oC$,the highest osmotic pressure is exhibited by $0.1 \ M$ solution of

Which of the following equimolar aqueous solutions has the highest freezing point?

The degree of dissociation of a $0.2 \, m$ aqueous solution of a weak acid $HX$ is $0.3$. If $K_f$ for water is $1.85 \, K \, kg \, mol^{-1}$,then the freezing point of the solution will be approximately ........... $^oC$.

In an aqueous solution of $NaCl$,the mole fraction of the solvent is $0.9$. Then,the relative lowering in vapour pressure will be ......... $atm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module