JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

A non-volatile, non-electrolyte solid solute when dissolved in $40$ g of a solvent, the vapour pressure of the solvent decreased from $760$ mm Hg to $750$ mm Hg. If the same solution boils at $320$ K, then the number of moles of the solvent present in the solution is _____. (Nearest integer)

[Given: boiling point of the pure solvent $= 319.5$ K, $K_b$ of the solvent $= 0.3$ K kg mol $^{-1}$ ]

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

From the relative lowering of vapour pressure, we have:

$\dfrac{P^0 - P_s}{P_s} = \dfrac{n_B}{n_A}$

where $n_B$ is the number of moles of solute and $n_A$ is the number of moles of solvent.

Substituting the given values:

$\dfrac{760 - 750}{750} = \dfrac{n_B}{n_A}$

$\dfrac{10}{750} = \dfrac{n_B}{n_A} \Rightarrow n_A = 75 n_B$

The elevation in boiling point is given by:

$\Delta T_b = T_b - T_b^0 = 320 - 319.5 = 0.5$ K

We also know that $\Delta T_b = K_b \times m$ , where $m$ is the molality of the solution.

$m = \dfrac{n_B}{\text{Mass of solvent in kg}} = \dfrac{n_B}{0.04}$

Substituting the values into the elevation in boiling point equation:

$0.5 = 0.3 \times \dfrac{n_B}{0.04}$

$n_B = \dfrac{0.5 \times 0.04}{0.3} = \dfrac{0.02}{0.3} = \dfrac{1}{15}$ mol

Now, substituting $n_B$ back to find the number of moles of the solvent $n_A$ :

$n_A = 75 \times \dfrac{1}{15} = 5$ mol

Answer: $5$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

Solution

From the relative lowering of vapour pressure, we have:

$\dfrac{P^0 - P_s}{P_s} = \dfrac{n_B}{n_A}$

where $n_B$ is the number of moles of solute and $n_A$ is the number of moles of solvent.

Substituting the given values:

$\dfrac{760 - 750}{750} = \dfrac{n_B}{n_A}$

$\dfrac{10}{750} = \dfrac{n_B}{n_A} \Rightarrow n_A = 75 n_B$

The elevation in boiling point is given by:

$\Delta T_b = T_b - T_b^0 = 320 - 319.5 = 0.5$ K

We also know that $\Delta T_b = K_b \times m$ , where $m$ is the molality of the solution.

$m = \dfrac{n_B}{\text{Mass of solvent in kg}} = \dfrac{n_B}{0.04}$

Substituting the values into the elevation in boiling point equation:

$0.5 = 0.3 \times \dfrac{n_B}{0.04}$

$n_B = \dfrac{0.5 \times 0.04}{0.3} = \dfrac{0.02}{0.3} = \dfrac{1}{15}$ mol

Now, substituting $n_B$ back to find the number of moles of the solvent $n_A$ :

$n_A = 75 \times \dfrac{1}{15} = 5$ mol

Answer: $5$

JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry