Question

The vapour pressure of pure liquid X at 27°C is 150 mm of Hg. If 0.5 mole of a non-volatile and non-electrolyte solute is dissolved in 50 g of X, then the vapour pressure of the solution will be(Molar mass of X = 25 g mol–1)

Answer 1

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Answer 2

The vapor pressure of the solution at $27^0C$ is 120 mm Hg

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 electrolytes)

$x_2$ = mole fraction of solute  =$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 0.5 moles of solute is present in 50 g of X, having molecular mass of 25g/mol

moles of solute= 0.5

moles of solvent (X) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50g}{25g/mol}=2moles$

Total moles = moles of solute + moles of solvent (X) = 0.5 + 2 = 2.5

$x_2$ = mole fraction of solute

=$\frac{0.5}{2.5}=$0.2

$\frac{150-p_s}{150}=1\times 0.2$

$p_s=120mmHg$

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