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A solution containing $0.5\;g$ of a solute (molar mass 130 g\;mol^{-1})$ in $50 \;g\;CCl_4$ yields a boiling point elevation of boiling point . The molar mass of solute is about

$\begin{array}{1 1} 85\;g\;mol^{-1} \\96\;g\;mol^{-1} \\ 106\;g\;mol^{-1} \\ 121\;g\;mol^{-1} \end{array} $

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Answer :$96\;g\;mol^{-1}$
$0.4\; K= K_b \large\frac{(0.5 g)/130 \;g \;mol^{-1})}{(50 \;g)}$ and $ 0.65 \;K= K_b \large\frac{(0.6 \;g) /M}{(50 \;g)}$
From these we get $\large\frac{0.4}{0.65} = \large\frac{0.5 /(130\;g\;mol^{-1})}{0.6 /M}$
=> $ M= \large\frac{0.4 \times 0.6 \times 130 }{0.65 \times 0.5}$$g\;mol^{-1} =96 \;g\; mol^{-1}$
answered Aug 7, 2014 by meena.p

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