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

12th Grade > Chemistry

SOLUTIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1. Which of the followings is correct –

For an non-ideal solution △Hmix > 0, △Vmix > 0.
In a solution showing negative deviation from Raoult's law A-B type of molecular interactions are stronger than A-A and B-B type of molecular interactions.
Nitric acid and Water solution shows positive deviation from Raoult's law.
Azeotrope boils at temperature below the boiling temperature of its constituents.

Discuss Question

Answer: Option B. -> In a solution showing negative deviation from Raoult's law A-B type of molecular interactions are stronger than A-A and B-B type of molecular interactions.
:
B

a) △ H_{m i x} > 0, △ V m i x > 0

aretrueformixturesshowing positivedeviations from Raoult'slaw only.Hence option (a)isincorrect
b) It is true
c) Nitric acid and water forms intermolecular H-bonding. Hence shows Negative deviation from Raoult's law.
d) Azeotropes are of two types – minimum and maximum boiling azeotropes. Only minimum boiling Azeotrope boils at temperature below the boiling temperature of its constituents. Hence option (d) is incorrect.

Question 2. Aquatic species feel more comfortable in winter than in summer because -

% of O2 in air decreases in summer due to high temperature
Atmospheric pressure decreases in summer leading to reduced solubility of gases in water
Solubility of gases in water decreases as the temperature increases due to increased Henry’s constant
Aquatic species feel skin burn due to increased temperature of the water.

Discuss Question

Answer: Option C. -> Solubility of gases in water decreases as the temperature increases due to increased Henry’s constant
:
C
According Henry’slaw ofsolubilityofgases -

P = K_{H} X

Withincreasein temperatureincreases,thus ‘x’ decreases given thepartialpressureof

O_{2}

remainssame.

Question 3. In an experiment on depression of freezing point, two beakers leveled as A and B were filled with 1000 g. of water each. 40 g and 80 g of a certain non-electrolytic solute were added to both the beakers and freezing point of the solutions were noted as

T_{f A}

and

T_{f B}

respectively. Now both the solutions were poured into an empty beaker C and the freezing point of the mixed solution is found to be

T_{f C}

. What is the ratio

T_{f A} : T_{f B} : T_{f C} ?

1 : 2 : 3
2 : 4 : 3
2 : 4 : 8
1 : 4 : 8

Discuss Question

Answer: Option B. -> 2 : 4 : 3
:
B

△ T_{f} = K_{f} m

\Rightarrow 0 - T_{f} = K_{f} m

\Rightarrow= - K_{f} m

If 40 g of solute contains 'x'~moles of it then 80 g of the solut ill contains '2x' molesofit

m_{A} = \frac{x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{x}{\frac{1000}{1000}} = x

m_{B} = \frac{2 x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{x}{\frac{1000}{1000}} = 2 x

m_{C} = \frac{3 x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{3 x}{\frac{1000}{1000}} = 1.5 x

Required ration

T_{f A} : T_{f B} : T_{f c} = 1.2 : 1.5 = 2 : 4 : 3

Question 4. Which of the following solutions in water will have the lowest vapour pressure

0.1 M,NaCl
0.1 M Na3 PO4
0.1 M, BaCl2
0.1 M, sucrose

Discuss Question

Answer: Option B. -> 0.1 M Na3 PO4
:
B
Depression in vapor pressure is a colligative property. More the no. of particles present in the solution lower will be the vapor pressure. consists of maximum ions hence it show lowest vapour pressure.

\begin{matrix} M o l e c u l e & N a C l & N a_{3} P o_{4} & B a C l_{2} & S u c r o s e N o . o f i o n s p e r m o l e c u l e & 1 + 1 = 2 & 3 + 1 = 4 & 1 + 2 = 3 & 1 \end{matrix}

Question 5. A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at 25⁰C. Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at 25⁰C. Calculate the molecular weight of the solute.

74.2
75.6
67.83
78.7

Discuss Question

Answer: Option C. -> 67.83
:
C

A Solution Containing 30 Gms Of Non-volatile Solute In Exact...

Question 6. A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at

20^{\circ}

is 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be

0.549
0.200
0.786
0.478

Discuss Question

Answer: Option D. -> 0.478
:
D
Mole fractionof pentane invapourphase =

Y_{1} = \frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{1}^{0} X_{1}}{P_{1}^{0} X_{1} + P_{2}^{0} X_{2}}

= \frac{440 \times \frac{1}{5}}{440 \times \frac{1}{5} + 120 \times \frac{1}{5}} = 0.478

Question 7. Which of the following liquid pairs shows a positive deviation from Raoult's law

Water-nitric acid
Benzene-methanol
Water-hydrochloric acid
Acetone-chloroform

Discuss Question

Answer: Option B. -> Benzene-methanol
:
B
(a) Water-Nitric acid solution shows negative deviation from Raoult's law due to hydrogen bonding
(b) Benzene-methonol solution shows positive deviation due to formation hydrogen bonding
(c) Water-HCl solution shows negative deviation due to formation of hydrogen bonding
(d) Acetone-Chloroform solutions shows negative deviation due to formation of hydrogen bonding
it is prudent to remember some of these examples of solutions showing positive and negative deviations from Rault's law along with examples of ideal solutions

\begin{matrix} N e g a t i v e d e v i a t i o n f r o m R a u l t^{'} s l a w & P o s i t i v e d e v i a t i o n f r o m R a u l t^{'} s l a w & i d e a l s o l u t i o n s ∙ C h l o r o f o r m - A c e t o n e ∙ W a t e r - N i t r i c a c i d ∙ W a t e r - H C l ∙ P h e n o l - A n i l i n e & ∙ B e n z e n e - M e t h y l a l c o h o l ∙ c a r b o n d i s u l f i e - A c e t o n e ∙ C h l o r o f o r m - e t h a n o l ∙ W a t e r - E t h y l a l c o h o l & ∙ B e n z e n e - T o l u e n e ∙ n - h e x a n e^{-} n - h e p t a n e ∙ E t h y l b r o m i d e - E t h y l i o d i d e ∙ C h l o r o B e n z e n e - b r o m o B e n z e n e \end{matrix}

Question 8. Assertion (A): Molality of solution does not change with change in temperature.
Reason (R): Molality is defined as no. of moles of solute per kg of solvent. Which of the following is correct?

Both A and R are true and R is correct explanation of A
Both A and R are true and R is not a correct explanation of A
A is true but R is false.
A is false but R is true.

Discuss Question

Answer: Option B. -> Both A and R are true and R is not a correct explanation of A
:
B
Molalitydoesnotchange with change in temperature, because it is defined as

\frac{n o . o f . 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}

. Neither no. of moles nor the wt. of solvent is changing with change in temperature. Hence (b) is the correct option.

Question 9. Ratio of Henry’s constant,

K_{H}

of two inert gases A

&

B is 2 at a given temperature. A mixture containing 4 moles of gas A and 8 moles of gas B is bubbled through water covered from top and solution is allowed to attain equilibrium. Assuming the absence of any other gases, what will be the ratio of mole fractions of gases in the solution.

Discuss Question

Answer: Option B. -> 8:9
:
B
Let ‘a’ and ‘b’ moles of gases dissolve in water while rest accumulates over the solution to create partial pressure.
From Henry’s Law –