AIIMS2019Chemistry-Solutions

AIIMS 2019 Chemistry Vapour Pressure MCQ Question

Type: MCQ-numerical-Medium-Class 12

Vapour pressure of CCl4\text{CCl}_4, at 25°C is 143 mm Hg. 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in 100 mL of CCl4\text{CCl}_4, Find the vapour pressure of the solution. (Density of CCl4,=1.58 g/cm3\text{Density of CCl}_4, = 1.58\text{ g/cm}^3)

A

141.93 mm Hg

B

94.39 mm Hg

C

199.34 mm Hg

D

143.99 mm Hg

Correct Answer

Option A

Detailed Explanation

Option A is correct because it accurately represents the relationship between the mole fraction of the solute (x₂) and the relative lowering of vapor pressure, given by the equation x2=P0PsP0x₂ = \frac{P₀ - Pₛ}{P₀}. In this equation, P0P₀ is the vapor pressure of the pure solvent (e.g., H₂O), and PsPₛ is the vapor pressure of the solution. The other options either provide incorrect numerical values or fail to correctly apply the formula, making them invalid in the context of this problem. Understanding this relationship is crucial for solving problems related to colligative properties in solutions.

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