JEE Main 2026 Apr 6 Shift 1, Chemistry Q6: Vapour Pressure Of A Solution

JEE Main2026Apr 6, Shift 1Chemistry

When $0.25$ moles of a non-volatile, non-ionizable solute was dissolved in $1$ mole of a solvent the vapor pressure of solution was $x %$ of vapor pressure of pure solvent. What is $x %$ ?

A
$50%$
B
$60%$
C
$70%$
D
$80%$

Solution

By Raoult's law for a non-volatile solute, $P_{s} = P^{0} x_{A}$ , where $x_{A}$ is the mole fraction of solvent.

$x_{A} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.80$ .

$∴ \frac{P _{s}}{P ^{0}} \times 100 = 80%$ .

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JEE Main 2026 Apr 6 Shift 1, Chemistry Q6: Vapour Pressure Of A Solution

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