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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
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1.00g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is $5.12 K kg mol^{−1}$. Find the molar mass of solute

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$W_2$ = 1.00g
$W_1$ = 50g
$\Delta T_f$ = 0.40 K
$K_f = 5.12 K kg mol^{−1}$
We know
$M_2=\large\frac{K_f\times W_2\times 1000}{W_1\times \Delta T_f}$
Substituting the values in the above equation, we get
$M_2=\large\frac{1000\times 5.12\times 1}{50\times 0.4}$
$\Rightarrow 256gmol^{-1}$
Hence (A) is the correct answer.
answered Jun 12, 2014 by sreemathi.v
 

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