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SubjectsChemistry
वाष्प दाब
In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how vapour pressure arises from the dynamic equilibrium between evaporation and condensation in a liquid. The topic explains the effect of temperature and the presence of a non-volatile solute, including lowering of vapour pressure. Students also connect vapour pressure with mole fraction through Raoult’s law and examine how ideal and non-ideal solutions differ, using equations and basic numerical applications.
TOPIC PRACTICE
Up to 5 questions from this page. Select your focus, then start.
Take 1 L of solution. Its mass is 1.04 × 1000 = 1040 g. The solution contains 0.5 mol glucose, whose mass is 0.5 × 180 = 90 g. Thus solvent mass = 1040 − 90 = 950 g = 0.950 kg. Molality = 0.5/0.950 = 0.526 mol kg⁻¹ approximately. Option B confuses molarity with molality, while option C correctly uses solvent mass.
The moles of water are n₁ = 9/18 = 0.50 mol. For the solute, x₂ = n₂/(n₁ + n₂) = 0.02. Solving gives 0.02(0.50 + n₂) = n₂, so n₂ = 0.010/0.98 ≈ 0.0102 mol. Therefore, molar mass = 0.18/0.0102 ≈ 17.6 g mol⁻¹, which is approximately 18 g mol⁻¹. Option A is correct.
For a dilute solution of a non-volatile solute, relative lowering of vapour pressure equals the solute mole fraction: x₂ = 0.05. The 18 g of water represents 18/18 = 1.00 mol. Let n₂ be solute moles. Then 0.05 = n₂/(1 + n₂), giving n₂ = 0.05/0.95 = 0.0526 mol. Hence, molar mass = 1/0.0526 ≈ 19 g mol⁻¹, so option B is correct.
Assume the solution contains a nonvolatile solute and use the mole-fraction definition xsolute = nsolute/(nsolute + nwater). The 18 g of water corresponds to 18/18 = 1.00 mol water. Therefore 0.04 = n/(1 + n), giving 0.04 + 0.04n = n and n = 0.04/0.96 = 0.04167 mol solute. Its molar mass is mass/moles = 1.8/0.04167 ≈ 43.2 g mol−1.
The usual relative-lowering-of-vapour-pressure method assumes a non-volatile solute, so the vapour pressure of the solution is attributed to the solvent alone. If the solute is volatile, it also enters the vapour phase and contributes its own partial pressure. The simple Raoult’s-law relation is then not applicable in the assumed form, causing an error in the molar-mass calculation unless both components’ vapour pressures are considered.
QUIZ COMPLETE
