0.5 m aqueous solution of a weak acid (HX) i | vedclass

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

Question

(B) The dissociation of the weak acid is represented as: $HX \rightleftharpoons H^{+} + X^{-}$.For the dissociation of an electrolyte,the van't Hoff f

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(B) The dissociation of the weak acid is represented as: $HX ightleftharpoons H^{+} + X^{-}$.For the dissociation of an electrolyte,the van't Hoff factor $i$ is given by $i = 1 + \alpha(n-1)$,where $n$ is the number of ions produced per molecule.Here,$n = 2$ and $\alpha = 0.20$ $(20 \%)$.So,$i = 1 + 0.20(2-1) = 1.2$.The lowering in freezing point is given by $\Delta T_{f} = i 	imes K_{f} 	imes m$.Substituting the values: $\Delta T_{f} = 1.2 	imes 1.86 	imes 0.5 = 1.116 \ K \approx 1.12 \ K$.

$0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20 \%$ ionized. If $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$,the lowering in freezing point of the solution is: (in $K$)

Similar Questions

Which of the following has the same van't Hoff factor $(i)$ as $K_3[Fe(CN)_6]$?

Which of the following salts has the same value of Van't Hoff factor $i$ as that of $K_4[Fe(CN)_6]$?

For the dissociation of $Na_2SO_4$,how is the degree of dissociation $\alpha$ used to calculate the van't Hoff factor $(i)$?

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

Two solutions of $KNO_3$ and $CH_3COOH$ are prepared separately. Molarity of both is $0.1 \ M$ and osmotic pressures are $P_1$ and $P_2$ respectively. The correct relationship between the osmotic pressures is :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module