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SubjectsChemistry
असामान्य आणविक द्रव्यमान
In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn why the experimentally determined molecular mass of a solute may differ from its expected value. The discussion connects abnormal molecular mass with the association or dissociation of solute particles in solution and explains how these changes affect colligative properties. Students also explore the van’t Hoff factor and use it to interpret and calculate corrected molar masses in solution-based problems.
TOPIC PRACTICE
Up to 7 questions from this page. Select your focus, then start.
An i value of 0.48 is far below unity, so the solution contains substantially fewer effective particles than expected. Strong association can produce this reduction; complete dimerisation has a limiting i of 0.50, so a value near it represents very strong association under the ideal model. Dissociation would increase i, and association raises rather than lowers apparent mass. Thus A is correct.
Starting from M_app = M/i, rearrange to M = iM_app. Substituting the data gives M = 1.25 × 96 = 120 g mol−1. Since i is greater than one, the true molar mass is larger than the apparent value, as expected for dissociation. Dividing instead of multiplying would give the incorrect option 76.8.
An i value of 5 means that one original solute unit gives five effective particles. Complete dissociation into five ions produces exactly this factor. Association would reduce i below one: complete pentamer, dimer, and other associations produce fewer groups than original particles. A normal undissociated solute has i = 1.
Dissociation breaks one solute unit into two or more particles, increasing the effective number of particles. Therefore, the van’t Hoff factor becomes greater than one, i > 1. Since M_observed = M_true/i, division by a number greater than one makes M_observed smaller than M_true. Thus option A is correct. Association produces the opposite trend, with i < 1 and a higher observed molar mass.
The van’t Hoff factor is the ratio of the actual effective number of particles to the number expected without association or dissociation. Thus, i = 0.0075 divided by 0.010 = 0.75. Because i is below 1, particles have combined and their number has decreased. This indicates association, such as dimerisation, not dissociation.
For abnormal molar-mass measurements, i = M true divided by M observed. Therefore, i = 90 divided by 45 = 2. A value greater than 1 means that the number of solute particles has increased, which is explained by dissociation into ions or smaller species. Association processes such as dimerisation or trimerisation decrease particle number and would produce i values below 1.
An AB electrolyte dissociates according to AB ⇌ A⁺ + B⁻ and produces two particles from one formula unit. For a fraction α dissociated, i = 1 + α because n = 2. Thus, 1.05 = 1 + α, giving α = 0.05, or 5% dissociation. This represents very small or slight dissociation. Complete dissociation would give i = 2, whereas association would make i less than 1.
QUIZ COMPLETE
