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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The vapour pressure of pure liquid solvent A is 0.80atm.When a non volatile substance B is added to the solvent,its vapour pressure drops to 0.40atm,the mole fraction of component B in the solution is

$\begin{array}{1 1}(a)\;0.25\\(b)\;0.75\\(c)\;0.50\\(d)\;0.40\end{array}$

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Answer: 0.5
In solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
This translates to the equation $\large\frac{p^0-p}{p^0}$$=x_2$, the expression on the left hand side is called the relative lowering of vapour pressure and is equal to the mole fraction of the solute.
$\large\frac{0.80-0.40}{0.80}$$=x_2 = 0.5$
answered Mar 25, 2014 by balaji.thirumalai
edited Jul 15, 2014 by balaji.thirumalai
 

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