MHTCET 16 May 2024 Shift 2 Paper

Concept:When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases. Relative lowering of vapour pressure is the fraction by which the vapour pressure of the pure solvent is lowered, expressed as $\frac{P^\circ - P}{P^\circ}$.Explanation:Step 1: Identify the given values.Vapour pressure of the pure solvent, $P^\circ = 32$ mm Hg.Vapour pressure of the solution, $P = 30$ mm Hg.The temperature is $25^\circ$C, which remains constant throughout the measurement.Step 2: Calculate the decrease in vapour pressure.Decrease in vapour pressure $= P^\circ - P = 32 - 30 = 2$ mm Hg.Step 3: Apply the formula for relative lowering of vapour pressure.Relative lowering of vapour pressure $= \frac{P^\circ - P}{P^\circ}$Substituting the values:$$\frac{P^\circ - P}{P^\circ} = \frac{32 - 30}{32} = \frac{2}{32} = 0.0625$$This is a dimensionless quantity, since both numerator and denominator have the same units (mm Hg).Answer:The relative lowering of vapour pressure is $0.0625$.Hence, the correct option is C. $0.0625$.