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NEET-UG and JEE Main Chemistry · Class 12

Abnormal molar mass and the van't Hoff factor — NEET & JEE Questions

Abnormal molar mass and the van't Hoff factor is a topic in Solutions, a Class 12 Chemistry chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 39 published NEET-UG and JEE Main questions on Abnormal molar mass and the van't Hoff factor, each with the marking scheme of the real paper.

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Abnormal molar mass and the van't Hoff factor MCQs with solutions

10 of the 39 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    The van't Hoff factor i is defined as the ratio of the observed colligative effect to the

    • A. colligative effect calculated for a nonelectrolyte at the same concentration
    • B. molar mass of the solvent
    • C. density of the solution
    • D. boiling point of the pure solvent
  2. Question 2 · difficulty L2 · understanding

    Match List I with List II List I List II A. van't Hoff factor, i I. Cryoscopic constant B. kf\mathrm{k_f} II. Isotonic solutions C. Solutions with same osmotic pressure III. Normal molar massAbnormal molar mass\mathrm{\frac{Normal\,molar\,mass}{Abnormal\,molar\,mass}} D. Azeotropes IV. Solutions with same composition of vapour above it Choose the correct answer from the options given below :

    • A. A-III, B-I, C-IV, D-II
    • B. A-III, B-I, C-II, D-IV
    • C. A-III, B-II, C-I, D-IV
    • D. A-I, B-III, C-II, D-IV
  3. Question 3 · difficulty L2 · understanding

    Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is

    • A. 1 : 4
    • B. 1 : 0.25
    • C. 1 : 0.20
    • D. 1 : 5
  4. Question 4 · difficulty L2 · understanding

    Which one of the following 0.06 M aqueous solutions has lowest freezing point?

    • A. Al 2 (SO 4 ) 3
    • B. C 6 H 12 O 6
    • C. KI
    • D. K 2 SO 4
  5. Question 5 · difficulty L2 · understanding

    If A 2 B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _______ × 10 −1 .

  6. Question 6 · difficulty L2 · understanding

    25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M AgNO3\mathrm{AgNO_3} solution when titrated using K2CrO4\mathrm{K_2CrO_4} as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer). (Given : Kf=2.0 K kg mol−1\mathrm{K_f=2.0~K~kg~mol^{-1}}) Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL−1^{-1}

  7. Question 7 · difficulty L2 · understanding

    Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (K b /K f ) is _________.

  8. Question 8 · difficulty L2 · understanding

    2 g of a non-volatile non-electrolyte solute is dissolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio xy{x \over y} (x : y). The value of y is ______________. (Nearest integer)

  9. Question 9 · difficulty L2 · understanding

    The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol −-1 in 200 mL of water at 27∘^\circC is ______________ Pa. [integer value] (use R = 0.083 L bar mol −-1 K −-1 )

  10. Question 10 · difficulty L2 · understanding

    Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C 2 H 5 OH is __________ (Integer answer) (i) 0.10 M Ba 3 (PO 4 ) 2 (ii) 0.10 M Na 2 SO 4 (iii) 0.10 M KCl (iv) 0.10 M Li 3 PO 4

Common mistakes in Abnormal molar mass and the van't Hoff factor

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Uses the inverse ratio and obtains 0.5.
  • Reverses the effect of association and dissociation on particle number.

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