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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
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The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg $mol^{–1}$

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$W_2 = 1.80g$
$W_1 = 90g$
$\Delta T_b$=354.11K-353.23K=0.88K
$K_b = 2.53$ K kg mol−1
We know,
$M_2=\large\frac{K_b\times W_2\times 1000}{W_1\times \Delta T_b}$
Substituting the values in the above equation, we get
$M_2=\large\frac{2.53\times 1.8\times 1000}{0.88\times 90}$
$\Rightarrow 58$
Therefore, molar mass of the solute, $M_2$ = 58 g $mol^{–1}$
Hence (C) is the correct answer.
answered Jun 11, 2014 by sreemathi.v
 

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