Q25 / 75 · JEE Main 2026 (04 April Shift 1) · Chemistry
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$^{-1}$]
[Given: boiling point of the pure solvent $= 319.5$ K, $K_b$ of the solvent $= 0.3$ K kg mol$^{-1}$]
Correct answer: 5
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: $5$
$\dfrac{P^0 - P_s}{P_s} = \dfrac{n_B}{n_A}$
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: $5$
Hence, the correct answer is 5.