JEE Main 2025 — Solutions Question with Solution

When a non-volatile solute is added to a solvent, it causes the vapour pressure of the solvent to decrease. In this scenario, when the vapour pressure decreases by 10 mm of Hg, the mole fraction of the solute in the solution is 0.2.

By understanding the relationship between vapour pressure change and mole fraction, we see that:

The change in vapour pressure ($P^{\circ }-P$) is directly proportional to the mole fraction of the solute ($X_{\text{solute}}$).

Therefore, if a 10 mm of Hg decrease corresponds to a mole fraction of 0.2, then a 20 mm of Hg decrease would correspond to a mole fraction of 0.4.

To find the mole fraction of the solvent ($X_{\text{solvent}}$), we use the formula:

$X_{\text{solvent}}=1-X_{\text{solute}}$

Substituting the value we found:

$X_{\text{solvent}}=1-0.4=0.6$

Thus, when the vapour pressure decreases by 20 mm of Hg, the mole fraction of the solvent is 0.6.