For a 0.1 , molal aqueous solution of 2-iodop | vedclass

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

Question

(B) The dissociation of $2-$iodopropanoic acid $(CH_3CH(I)COOH)$ is given by: $CH_3CH(I)COOH \rightleftharpoons CH_3CH(I)COO^- + H^+$. For a weak acid

Vedclass · Accepted Answer

$-0.1953$

Vedclass · Answer

(B) The dissociation of $2-$iodopropanoic acid $(CH_3CH(I)COOH)$ is given by: $CH_3CH(I)COOH ightleftharpoons CH_3CH(I)COO^- + H^+$. For a weak acid,the van't Hoff factor $i = 1 + \alpha(n-1)$. Here,$n = 2$ and $\alpha = 0.05$ $(5 \%)$. So,$i = 1 + 0.05(2-1) = 1.05$. The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Given $K_f = 1.86 \, K \, kg \, mol^{-1}$ and $m = 0.1 \, molal$. $\Delta T_f = 1.05 	imes 1.86 	imes 0.1 = 0.1953 \, K$. The freezing point of the solution is $T_f = T_f^0 - \Delta T_f = 0 - 0.1953 = -0.1953 \, ^oC$.

For a $0.1 \, molal$ aqueous solution of $2-$iodopropanoic acid,the degree of dissociation is $5 \%$. The freezing point of the solution will be ............ $^o C$.

Similar Questions

The boiling point (in ${}^{\circ}C$) of $0.1 \, molal$ aqueous solution of $CuSO_{4} \cdot 5H_{2}O$ at $1 \, bar$ is closest to $.....$
[Given : Ebullioscopic (molal boiling point elevation) constant of water,$K_{b}=0.512 \, K \, kg \, mol^{-1}$ ]

Which of the following has the highest boiling point at $1 \, atm$ pressure?

Which of the following is a non-electrolyte?

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

The freezing point of a solution prepared by adding $8.1 \ g$ of $HBr$ ($90\%$ ionized) to $100 \ g$ of water is $...... \ ^\circ C$. (Given $K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module