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Lowering of vapour pressure due to a solute in 1 molal aqueous solutions at $100^{\large\circ}C$ is

$\begin{array}{1 1}(a)\;352torr\\(b)\;312torr\\(c)\;13.44torr\\(d)\;14.12torr\end{array}$

1 Answer

Relative lowering of V.P
$\large\frac{P^0-P}{P^0}$$=X_i$=mole fraction
$P^0-P=\Delta P=X_i\times P^0$
$\Rightarrow \big(\large\frac{1}{1+\Large\frac{1000}{18}}\big)$$\times 760$
$\Rightarrow \big(\large\frac{1}{1+55.55}\big)$$\times 760$
$\Rightarrow 0.0176\times 760$
$\Rightarrow 13.44$
Hence (c) is the correct answer.
answered Feb 14, 2014 by sreemathi.v

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