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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} $

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1 Answer

$W_1=20.2g$
$W_2=0.2g$
$\Delta T_f=0.45^{\large\circ}C$
$K_f=5.12Kkg mol^{-1}$
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.
answered Aug 8, 2014 by sreemathi.v
 

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