JEE Main 2026 Apr 4 Shift 1, Chemistry Q25: Colligative Properties
JEE Main2026Apr 4, Shift 1Chemistry
Q.
A non-volatile, non-electrolyte solid solute when dissolved in g of a solvent, the vapour pressure of the solvent decreased from mm Hg to mm Hg. If the same solution boils at K, then the number of moles of the solvent present in the solution is _____. (Nearest integer)
[Given : boiling point of the pure solvent K, of the solvent K kg mol]
Correct answer: 5
Solution
$\dfrac{P^\circ-P_s}{P_s} = i\cdot\text{molality} \times \dfrac{M_{\text{solvent}}}{1000}$
$\Delta T_b = i\cdot K_b\cdot\text{molality} \Rightarrow \text{molality} = \dfrac{0.5}{0.3}$
$\text{Molecular Mass} = \dfrac{600}{75}\,\text{g}$
$\text{Moles} = \dfrac{40}{600/75} = 5$
Concept behind this question
Colligative PropertiesNotes, formulas and examples →More previous year questions on Colligative Properties
Practice more Chemistry
Concept-wise practice with instant solutions on Fundamenthol.
Start practicing →