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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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A sucrose solution in 50 g water boils at $100.15^{\circ}C$ , If this solution is heated till its temperature is $100.375^{\circ}C$, how much of water is evaporated ? Given :$ K_b(water)=0.52\;K\;kg\;mol^{-1}$

$\begin{array}{1 1} 30\;g \\25\;g \\ 15\;g \\10\;g \end{array} $

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1 Answer

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Answer : 30 g
Since $\Delta T_b=K_b m =K_b \bigg(\large\frac{n_2}{m_1}\bigg)$
We have $0.15 K=K_b \bigg( \large\frac{n_2}{0.05\;kg}\bigg)$
and $ 0.375 K= K_b \bigg(\large\frac{n_2}{m_1}\bigg) $
Thus $\large\frac{0.15 K}{0.375\;K}=\frac{m_1}{0.05 \;Kg}$
or $m_1= \bigg( \large\frac{0.15\;K}{0.375\;K}\bigg)$$(0.05\;Kg)=0.02\;kg$
Mass of water evaporated $=50 g =30 g$
answered Aug 7, 2014 by meena.p
 

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