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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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The boiling point elevation of 0.30g acetic acid in 100g benzene is 0.0633K.Calculate the molar mass of acetic acid from this data?What conclusion can you draw about the molecular state of the solute in the solution?[Given $K_b$ for benzene = $2.53Kkgmol^{-1}]$

$\begin{array}{1 1}119.90\\129.90\\139.90\\149.90\end{array} $

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Here
$W_1=100g$
$W_2=0.30g$
$\Delta T_b=0.0633K$
$K_b=2.53Kkg mol^{-1}$
$\therefore$ Molar mass of acetic acid $M_2=\large\frac{K_b\times W_2\times 1000}{\Delta T_b \times W_1}$
$\Rightarrow \large\frac{2.53\times 0.30\times 1000}{0.0633\times 100}$
$\Rightarrow 119.90$
Van't Hoff factor
$i=\large\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$
$\;\;=\large\frac{60}{119.90}$
$\;\;=0.5$
Hence $i < 1$ ,therefore the solute (acetic acid) is associated.
answered Aug 7, 2014 by sreemathi.v
 

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