Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
0 votes

The freezing point of a solution containing 0.2g of acetic acid is 20.0g of benzene is lowered by $0.45^{\large\circ}$C.Calculate Van't Hoff factor .[For benzene,$K_f=5.12Kkg mol^{-1}]$.What conclusion can you draw from the value of Van't Hoff factor obtained.

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

Can you answer this question?

1 Answer

0 votes
$\Delta T_f=0.45^{\large\circ}C$
$K_f=5.12Kkg mol^{-1}$
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
$i=\large\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$
Here,i < 1.Therefore the non-volatile solute (acetic acid) is associated.
answered Aug 8, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App