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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The vapour pressure of a solvent is found to be $400 mm$ Hg at $300 K$ When a certain amount of nonvolatile solute added to $50\;mol$ of the solvent its vapour pressure decreases to $360\;mm Hg$. The amount of solute added is:

$\begin{array}{1 1} 2.41 \;mol \\3.42 \; mol \\4.54\;mol\\5.45\; mol\end{array} $

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Answer: $4.54\;mol$
$- \Delta p = (400 -360) mmHg=40 mmHg$
$x_2 = \large\frac{- \Delta p }{p \hat i }$
$\qquad= \large\frac{40 mm Hg}{400 mm Hg} $$=0.1$
Thus $x_2 = \large\frac{n_2}{n_1+n_2}$
$\qquad= \large\frac{n_2}{500 \;mol +n_2}$$=0.1$
(or) $n_2 =\large\frac{5 mol}{1.1} $$=4.54 \;mol$
answered Aug 5, 2014 by meena.p
 

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