Solutions(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

SOLUTIONS MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21.
Given that

△ T_{f}

if is the diperssion in freezing point of the solvent of non-volatile solute of molality m, the quantity

l i m m \to 0 (\frac{△ T_{f}}{m}) i s e q u a l t o

Zero
One
∞
Finite value

Discuss Question

Answer: Option D. -> Finite value
:
D

△ T_{f} = K_{f} m

l i m m \to 0 (\frac{△ T_{f}}{m}) = l i m m \to 0 (K_{f}) = K_{f}

Question 22. A solution of urea contains 8.6 gm/litre (mol. Wt. 60.0). It is isotonic with a 5% (W/W) solution of a non-volatile solute having density 1 g/cc. The molecular weight of the solute will be

348.8
34.88
3488
861.2

Discuss Question

Answer: Option A. -> 348.8
:
A
Isotonic solutionshavesameosmoticpressure
As Osmotic pressure = C R T

\Rightarrow C_{1} = C_{2}

\Rightarrow \frac{n_{1}}{V_{1}} = \frac{n_{2}}{V_{2}}

5 %

W/W solutionmeans 100g ofsolution contains 5gofsolute

∴ \frac{\frac{8.6}{60}}{1} = \frac{\frac{5}{M}}{\frac{100}{1000}}

\Rightarrow M = \frac{5 \times 60 \times 10}{8.6} = 348.8

Question 23. One litre of 1 molar

H_{2} S O_{4}

solution has been diluted by 5 litre water, the normality of the solution is

0.2N
5N
10N
0.33N

Discuss Question

Answer: Option D. -> 0.33N
:
D

M_{1} V_{1} = M_{2} V_{2}

w h e r e V_{1} = 1 l i t r e, M_{1} = 1 m

V_{2} = 1 + 5 = 6 l i t r e, M_{2} = ?

thus,

1 \times 1 = 6 \times M_{2}

M_{2} = \frac{1}{6}

[^{'} n^{'} f a c t o r o f H_{2} S O_{4} = 2]

But normality='n' factor

\times

molarity = 2

\times

\frac{1}{6} = \frac{1}{3}

N = \frac{1}{3}

Question 24. Solubility of a gas X in liquid Y is given by Henry's law written as

P = K_{H} \times x

. Where

K_{H}

is Henry constant, P is partial pressure of the gas and x is mole fraction of gas in liquid. Which of the following is incorrect about

K_{H} -

It depends on nature of gas X
It is independent of nature of liquid Y
It is a function of Temperature
It changes with change in partial pressure of gas.

Discuss Question

Answer: Option D. -> It changes with change in partial pressure of gas.
:
D
Henry constant

K_{H}

dependsonnatureofgas, solvent andtemperature.But itdoes notchangewith partialpressureof thegas.

Question 25. Benzene and toluene form nearly ideal solutions. At

20^{\circ}

, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at

20^{\circ}

for a solution containing 78g of benzene and 46g of toluene in torr is

50
25
37.5
53.5

Discuss Question

Answer: Option A. -> 50
:
A
AccordingtoRault'slaw-

P_{B} = P_{B}^{0} \times X_{B} = 75 \times \frac{\frac{78}{78}}{\frac{78}{78} + \frac{46}{92}} = 75 \times \frac{1}{1 + 0.5} = 50 t o r r

Question 26. Shikhar was to perform a neutralization reaction in laboratory. In order to neutralize 500 ml of

2 M N a O H

solution he was supposed to add certain volume of

1 M H_{3} P O_{3}

solution, but by mistake, he added the same volume of water instead. What is the molality of the resulting solution – Assume density of the original solution =

1.08 g / c c

Density of water = 1 g/cc.

0.67
1
1.2
2

Discuss Question

Answer: Option B. -> 1
:
B
Wt. of original solution = Density x volume =

1.08 g / c c \times 500 m l = 540 g

Wt. of NaOH + = No. of moles x molecular Wt. =

\frac{500 \times 2}{1000} \times 40 g = 40 g

Wt. of solvent i.e water in original solution = 540-40 = 500 g
if

V_{2}

be the volume of

H_{3} P O_{3}

required for neutralizing the solution then-

M_{1} V_{1} = M_{2} V_{2}

\Rightarrow V_{2} = \frac{M_{1} V_{1}}{M_{2}} = \frac{2 \times 500}{1 \times 2} m l = 500 m l

Note-Basicity of

H_{3} P O_{3} = 2

Wt. of water added = Density of water

\times

Volume =

\frac{1 g}{c c} \times 500 m l = 500 g

Total Wt. of solvent i.e water = 500 + 500 = 1000g
Molality of solution =

\frac{M o l e s o f s o l u t e}{W t . o f s o l v e n t i n k g} = \frac{40 / 40}{1000 / 1000} = 1 m o l e / k g

Hence Optin-(b) is the correct Answer.

Question 27. Which one of the following would produce maximum elevation in boiling point?

0.1 M glucose
0.2 M glucose
0.1 M barium chloride
0.1 M magnesium sulphate

Discuss Question

Answer: Option C. -> 0.1 M barium chloride
:
C
BaCl₂gives maximum number ofions. Hence, its boiling point is maximum.

Question 28. Which one of the following mixtures can be separated into pure components by fractional distillation

Benzene – toluene
Water – ethyl alcohol
Water – nitric acid
Water – hydrochloric acid

Discuss Question

Answer: Option A. -> Benzene – toluene
:
A
Only Benzene-Toluene mixture forms an ideal solution. Rest shows either positive or negative deviation from Raoult's law hence will form Azeotrope which can't be separated by fractional distillation.

Question 29. Unit of mole fraction is