Question

A liquid has vapour pressure 35 x 10^3 Nm^-2 at 298 K. If the solution
contains 0.2 mole fraction of a solute, the vapour pressure of the
solution will be​

Answer 1

Answer:

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Explanation:

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Answer 2

Vapour pressure of the  solution will be​ $28\times 10^3Nm^{-2}$

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolyte)  

$x_2$ = mole fraction of solute = 0.2  

$\frac{35\times 10^3-p_s}{35\times 10^3}=1\times 0.2$

$p_s=28\times 10^3Nm^{-2}$

Thus the vapour pressure of the  solution will be​ $28\times 10^3Nm^{-2}$

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