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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The elevation in boiling point when $20.16\;g$ of freshly prepared $CuCl_2$ (relative molar mass : $134.4$) is dissolved in $1\;kg$ of water $(K_b=0.52\;kg\;K \;mol^{-1})$ is

$\begin{array}{1 1} 0.075^{\circ} C \\ 0.150^{\circ}C \\ 0.234^{\circ}C \\0.468^{\circ}C \end{array} $

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Answer : $0.234^{\circ}C$
Molality of given solution is $ m= \large\frac{n_2}{m_1}$
$\qquad = \large\frac{m_2/M_2}{m_1} $
$\qquad= \large\frac{(20.16\;g)/(134.4 \;g \;mol)}{1\;kg}$
$\qquad= 0.15\;mol\;kg^{-1}$
Molality of ions in solution is $3 m$ that is $0.45\;mol\;kg^{-1}$
$\Delta T_b =K_b$
$m= (0.52 \;kg\;K\;mol^{-1}) (0.45 \;mol\;kg^{-1})$
$\qquad= 0.234\;K$
$\qquad= 0.234^{\circ}C$
answered Aug 6, 2014 by meena.p
 

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