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Test: Abnormal Molar Masses (NCERT) - NEET MCQ


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Test: Abnormal Molar Masses (NCERT) - Question 1

Why is the molecular mass determined by measuring colligative property in case of some solutes is abnormal?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 1

Due to association or dissociation of solute molecules there is a change in number of particles. Since colligative properties depend on number of particles there is a change in molecular mass.

Test: Abnormal Molar Masses (NCERT) - Question 2

Which of the following representations of i (van't Hoff factor) is not correct?

Test: Abnormal Molar Masses (NCERT) - Question 3

Which of the following relations is not correctly matched with the formula?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 3

In case of dissociation: An → nA
Initial number of moles: 1  0
After dissociation:  1 − α nα
No. of particles = 1 − α + nα

Test: Abnormal Molar Masses (NCERT) - Question 4

Which of the following will have same value of vant's Hoff factor as that of K4[Fe(CN)6]?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 4

K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4-
Al2​(SO4​)3​ → 2Al3+ + 3SO42-

Test: Abnormal Molar Masses (NCERT) - Question 5

Arrange the following aqueous solutions in the order of their increasing boiling points
(i) 10−4M NaCl
(ii) 10−4M Urea
(iii) 10−3M MgCl2
(iv) 10−2M NaCl

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 5

10−4M NaCl i = 2
10−4M Urea i = 1
10−3M MgCl2 i = 3
10−2M NaCl i = 2
More the value of i, C, more will be the elevation in boiling point hence increasing order of boding point is 10−4M Urea < 10−4M NaCl < 10−3M MgCl2 < 10−2M NaCl

Test: Abnormal Molar Masses (NCERT) - Question 6

Which of the following has the highest freezing point?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 6

C6H12O6 is a non-electrolyte hence furnishes minimum number of particles and will have maximum freezing point.
ΔTf = iKfm or ΔTf ∝ i and ΔTf = Tf − Tf

Test: Abnormal Molar Masses (NCERT) - Question 7

If α is the degree of dissociation of Na2SO4, the vant Hoff's factor (i) used for calculating the molecular mass is:

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 7

Test: Abnormal Molar Masses (NCERT) - Question 8

For which of the following solutes the van’t Hoff factor is not greater than one?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 8

Urea is non-electrolyte, hence will not dissociate to give ions.

Test: Abnormal Molar Masses (NCERT) - Question 9

What will be the degree of dissociation of 0.1M Mg(NO3)2 ​solution if van't Hoff factor is 2.74?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 9

Mg(NO3)2 → Mg2+ + 2NO3

Degree of dissociation = 0.87 x 100 = 87%

Test: Abnormal Molar Masses (NCERT) - Question 10

Which of the following will have the highest f.pt. at one atmosphere?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 10

For the same concentration of different solvents any colligative property ∝ = i
For NaCl, i = 2
Sugar solution, i = 1
BaCl2, i = 3; FeCl3, i = 4
Thus, for sugar solution depression in freezing point is minimum i.e., highest freezing point.

Test: Abnormal Molar Masses (NCERT) - Question 11

A solute X when dissolved in a solvent associates to form a pentamer. The value of van't Hoff factor (i) for the solute will be

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 11

5A → A5

Test: Abnormal Molar Masses (NCERT) - Question 12

What will be the freezing point of a 0.5m KCl solution? The molal freezing point constant of water is 1.86C m−1.

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 12

ΔTf = iKf × m = 2 × 1.86 × 0.5 = 1.86∘ C
Tf = Tf − ΔTf = 0 − 1.86 = −1.86∘ C

Test: Abnormal Molar Masses (NCERT) - Question 13

What amount of CaCl2 (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5atm at 27oC?

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 13


Amount of CaCl2 = n × M = 0.0164 × 111 = 1.820 g

Test: Abnormal Molar Masses (NCERT) - Question 14

The van't Hoff factor of a 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 14

KCl ⇔ K+ Cl  (n = 2)

Test: Abnormal Molar Masses (NCERT) - Question 15

The elevation in boiling point of a solution of 9.43g of MgCl2 in 1kg1kg of water is (Kb = 0.52 K kg mol−1, Molar mass of MgCl2 = 94.3 g mol−1)

Detailed Solution for Test: Abnormal Molar Masses (NCERT) - Question 15

MgCl2 ⇔ Mg2+ + 2Cl, i = 3

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