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The elevation of boiling point of water produced by dissolving $1.17 g$ sodium chloride in $100 g$ water $K_b=0.512\;K\;kg\;mol^{-1})$ is

$\begin{array}{1 1} 0.103\;K \\ 0.205\;K \\ 0.309\;K \\0.410\;K \end{array} $

1 Answer

Answer : $0.205\;K$
Molality of sodium chloride $= \large\frac{(1.17/58.5)mol}{0.1 \;kg} $$=0.2 \;mol\;kg^{-1}$
Molality of ions in solution $=2 \times 0.2 \;mol\;kg^{-1}$.
$\Delta T_b =K_bm =(0.512 ) (2 \times 0.2 )K =0.205\;K$
answered Aug 6, 2014 by meena.p
 

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