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JEE Exam  >  JEE Test  >  Chemistry Main & Advanced  >  Test: Van't Hoff Factor (Liquid Solution) - JEE MCQ

Van't Hoff Factor (Liquid Solution) - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Van't Hoff Factor (Liquid Solution) (20 Questions)

You can prepare effectively for JEE Chemistry for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Van't Hoff Factor (Liquid Solution)". These 20 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 20

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Test: Van't Hoff Factor (Liquid Solution) - Question 1
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Only One Option Correct Type
This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

Q.

The degree of dissociation (α ) o f weak electrolyte AxBy is related to van’t Hoff factor (i) by the expression

[AIEEE 2011]

Detailed Solution: Question 1

 (a) van’t Hoff factor (i) = [1 + (n - 1)α] where, n = number of ions from one mole of solute

α = degree of ionisation/association 

Test: Van't Hoff Factor (Liquid Solution) - Question 2
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0.004 M Na2S04 aqueous solution is isotonic with 0.01 M glucose solution at 300 K. Thus, degree of dissociation of Na2S04 is

Detailed Solution: Question 2

i (van’t Hoff factor) = 1 + (y — 1) x = (1 + 2 x)

∴ π, (osmotic pressure) = MRTi = 0.004 x R x 300 x (1 + 2x)

Glucose is a non-electrolyte.
Hence,

i = 1

π2 = mRTi = 0.01 x R x 300 x 1 Two solutions are isotonic, hence

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Test: Van't Hoff Factor (Liquid Solution) - Question 3
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When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene, a freezing point depression of 2 K is observed. [Kf (benzene) = 1.72Kmol-1 kg]. The van’t Hoff factor (i) is 

[IIT - JEE 2007]

Detailed Solution: Question 3

w1| = 20.0 g

m1 = 11x 12 + 8 x 1 + 2 x 16 = 172 g mol-1  w2 = 50.0 g 

Test: Van't Hoff Factor (Liquid Solution) - Question 4
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An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb /Kf = 0.3, then AB

Detailed Solution: Question 4

Thus, AS behaves as a non-electrolyte at the freezing point of solution (remains non-ionised).

Test: Van't Hoff Factor (Liquid Solution) - Question 5
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Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)

Detailed Solution: Question 5

 Osmotic pressure is a colligative property, hence osmolarity is the molarity of the solution involving van't Hoff factor

Thus, osmolarity = 0.02 x i= 0.02 x 5 = 0.I0 osmol

Test: Van't Hoff Factor (Liquid Solution) - Question 6
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AB and CD are two non -volatile solutes dissolved in same solvent, showing following behaviour

AB is ionised in solvent, α = degree of ionisation

CD is dimerised in solvent, 3 = degree of dimerisation

If α = 0.80, then β is

Detailed Solution: Question 6

  • Calculating Van't Hoff Factor for AB:

    • AB dissociates into A⁺ and B⁻.
    • Initial moles of AB = 1
    • Moles dissociated = α
    • Moles of AB remaining = 1 - α
    • Moles of A⁺ formed = α
    • Moles of B⁻ formed = α
    • Total moles after dissociation = (1 - α) + α + α = 1 + α
    • Van't Hoff factor (i_AB) = (1 + α) / 1 = 1 + α

  • Calculating Van't Hoff Factor for CD:

    • CD dimerizes into (CD)₂.
    • Initial moles of CD = 1
    • Moles dimerized = β
    • Moles of CD remaining = 1 - β
    • Moles of (CD)₂ formed = β / 2
    • Total moles after dimerization = (1 - β) + (β / 2) = 1 - β/2
    • Van't Hoff factor (i_CD) = (1 - β/2) / 1 = 1 - β/2

  • Boiling Point Elevation Equation:

    • ΔTb = i * Kb * m (where Kb is the ebullioscopic constant and m is molality)

  • Given Relationship:

    • ΔTb(AB) = 2 * ΔTb(CD)
    • i_AB * Kb * m = 2 * i_CD * Kb * m
    • Since Kb and m are the same for both solutions, we can cancel them out:
    • i_AB = 2 * i_CD

  • Substituting the Van't Hoff Factors:

    • 1 + α = 2 * (1 - β/2)

  • Solving for β:

    • We are given α = 0.80
    • 1 + 0.80 = 2 * (1 - β/2)
    • 1.80 = 2 - β
    • β = 2 - 1.80
    • β = 0.20

Test: Van't Hoff Factor (Liquid Solution) - Question 7
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Which has the highest boiling point? 

Detailed Solution: Question 7

Tb (boiling point) = T0(solvent) + ΔTb

ΔTb = m x Kb x i 

m(same) = 0.1 molal

Δ Tb α i

Greater the value of i, larger the boiling point of solution.

*Multiple options can be correct
Test: Van't Hoff Factor (Liquid Solution) - Question 8
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One or More than One Options Correct Type
This section contains 3 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

Following solutions have been provided at temperature T K.
I. 1M aqueous glucose solution.
II. 1M aqueous sodium chloride solution.
III. 1M aqueous ammonium phosphate solution.
IV. 1M benzoic acid in benzene.
Select correct statements for the above solutions.

Detailed Solution: Question 8

In benzene, benzoic acid forms dimer by intermolecular H-bonding.

Thus, osmotic pressure

Thus, solutions are not isotonic, (a) is incorrect.
III has maximum osmotic pressure, thus hypertonic of l, II and IV, (b) is correct.
IV has minimum osmotic pressure, thus hypotonic of I, II and III. Thus, (c) is correct.
II is hypotonic of III but hypertonic of I and IV, thus (d) Is correct.

*Multiple options can be correct
Test: Van't Hoff Factor (Liquid Solution) - Question 9
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Which of the following statements are correct?

Detailed Solution: Question 9

(a,c)

(a) Vapour pressure of a component in a solution,  by Raoult's law, thus correct.

(b) π = i MRT

For NaBr,

y = 2, / = 1 + (y - 1)x = 2

M = 1, T = (273 + t )

∴  π = 2R (273 + t) thus, incorrect.

(c)

π = i MRT

π  ∝ for equimolar solution

i = 1 for sucrose

i = (1 + x)forCH3COOH (x < < 1)

for CH3COOH being weak acid

i = 2 for KCI

i = 3 for K2S04

Thus, π (osmotic pressure) of 0.01 M solutions of sucrose < CH3COOH < KCI < K2S04 Thus, correct.

 and K1 are different, hence ΔTf will be different, thus incorrect.

*Multiple options can be correct
Test: Van't Hoff Factor (Liquid Solution) - Question 10
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Matching List Type
Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

Q.

Match the different solutions in Column I with their ΔTb in Column II and select the answer from the codes given below.

Detailed Solution: Question 10

Thus, (i) → (p),   (ii) → (r), (iii) → (s), (iv)→(t), (v) → (r)

*Multiple options can be correct
Test: Van't Hoff Factor (Liquid Solution) - Question 11
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In a 0.2 molal aqueous solution of a weak acid (HX), depression in freezing point is 0.383° Kf is 1.86° mol-1 kg. Assum e molarity equal to molality.
Match the parameters in Column I with their values in Column II and select the answer from the codes given below.

Detailed Solution: Question 11

(d) Depression in freezing poin t is a colligative property.

Thus, (i) -> (r) (ii) -+ (p) (iii) -4 (s) (iv) -> (q)

Test: Van't Hoff Factor (Liquid Solution) - Question 12
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Comprehension Type
This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage I

Hgl2 is insoluble in water. This solute is added into Kl solution.

Q.

In the above case,

Detailed Solution: Question 12

Test: Van't Hoff Factor (Liquid Solution) - Question 13
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Hgl2 is insoluble in water. This solute is added into Kl solution.

Q.

Passage I van’t Hoff factor in this example

Detailed Solution: Question 13

Test: Van't Hoff Factor (Liquid Solution) - Question 14
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Passage II

A solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1 kg]

Q.

van’t Hoff factor (/') of the dibasic acid is

Detailed Solution: Question 14

y = 3 = number of ions from one unit

i = van't Hoff factor = 1 + (y - 1)x = (1 + 2x)

Also, 10 mL of dibasic acid = 12mL of 0.1 N NaOH

N1 (dibasic acid) = 0.12 g equivL-1 

0.12 x equivalent weight = concentration = 10 g L-1'

Test: Van't Hoff Factor (Liquid Solution) - Question 15
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A solution containing 10 g of a dibasic acid in 1000 g of water freezes at -0.15° C. 10 mL of the acid is neutralised by 12 mL of 0.1 N NaOH solution. [Kf (H20 ) = 1.86° mol-1 kg]

Q.

pH of the solution of the acid is

Detailed Solution: Question 15

y = 3 = number of ions from one unit

i = van't Hoff factor = 1 + (y - 1)x = (1 + 2x)

Also, 10 mL of dibasic acid = 12mL of 0.1 N NaOH

N1 (dibasic acid) = 0.12 g equivL-1 

0.12 x equivalent weight = concentration = 10 g L-1'

Test: Van't Hoff Factor (Liquid Solution) - Question 16
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Passage III

A solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.

Q.

Apparent molecular weight of NaCI is 

Detailed Solution: Question 16

From ΔT, of cane sugar, K, can be determined. 

For NaCI solution,

Test: Van't Hoff Factor (Liquid Solution) - Question 17
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Passage III

A solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.

Q.

van’t Hoff factor (i) is

Detailed Solution: Question 17

From ΔT, of cane sugar, K, can be determined. 

For NaCI solution,

*Answer can only contain numeric values
Test: Van't Hoff Factor (Liquid Solution) - Question 18
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Direction:

This section contains 5 questions. When worked out will result in an integer from 0 to 9 {both inclusive).

Q.

Relative decrease in vapour pressure of an aqueous NaCI solution is 0.167. Thus, number of moles of NaCI present in 180 g of H20 is ... .


Detailed Solution: Question 18

*Answer can only contain numeric values
Test: Van't Hoff Factor (Liquid Solution) - Question 19
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Aqueous solution of barium phosphate which is 100% ionised has ΔTf / Kf as 0.40. Hence, given solution is x * 10-2 molal. What is the value of x?


Detailed Solution: Question 19

(8) Barium phosphate is Ba3( P04)2.

Test: Van't Hoff Factor (Liquid Solution) - Question 20
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0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI freezes at - 0.00732° C. [Kf (H20) = 1.86° mol-1kg]. How many moles of ions does 1.0 mole of the salt produce on being dissolved in water?

Detailed Solution: Question 20




Total = 2 ions.
1 mole of the salt produces 2 moles of ions

Hence, option(A) is correct

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This test contains 20 MCQs on Van't Hoff Factor (Liquid Solution) with detailed solutions for each question. It is part of the Chemistry for JEE Main & Advanced course on EduRev, designed to align with the current JEE syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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