$\begin{array}{1 1}0.52\\0.62\\0.72\\0.42\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$W_1=20.2g$

$W_2=0.2g$

$\Delta T_f=0.45^{\large\circ}C$

$K_f=5.12Kkg mol^{-1}$

Molar mass of acetic acid $M_2=\large\frac{K_f \times W_2\times 1000}{\Delta T_f\times W_1}$

$\Rightarrow \large\frac{5.12\times 0.2 \times 1000}{0.45\times 20.0}$

$\Rightarrow 113.77gmol^{-1}$

Actual molar mass of acetic acid =$60gmol^{-1}$

Van't Hoff factor

$i=\large\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$

$i=\large\frac{6}{113.77}$

$\;\;=0.52$

Here,i < 1.Therefore the non-volatile solute (acetic acid) is associated.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...