JEE Main 2026 — Solutions Question with Solution

From the relative lowering of vapour pressure, we have:

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

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

Substituting the given values:

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

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

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.

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

Substituting the values into the elevation in boiling point equation:

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

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

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

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

Answer: $5$