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

12th Grade > Chemistry

SOLUTIONS MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. The molecular mass of benzoic acid in benzene, as determined by depression in freezing point method, corresponds to:

ionization of benzoic acid
dimerization of benzoic acid
trimerization of benzoic acid
solvation of benzoic acid

Discuss Question

Answer: Option B. -> dimerization of benzoic acid
:
B
Benzoic acid undergoes association in benzene i.e. it dimerizes.

Question 12. Boiling point of chloroform was raised by 0.323 K, when 0.5143 g of anthracene was dissolved in 35 g of chloroform. Molecular mass of anthracene is(

= 3.9 kg mol^-1)

79.42 g/mol
132.32 g/mol
177.42 g/mol
242.32 g/mol

Discuss Question

Answer: Option C. -> 177.42 g/mol
:
C

Boiling Point Of Chloroform Was Raised By 0.323 K, When 0.51...

Question 13. Two isotonic solutions must have ----

Same solute and solvent
Same solute but different solvent
Same solvent but different solute
Both Solute and solvent may be different.

Discuss Question

Answer: Option D. -> Both Solute and solvent may be different.
:
D
For two solutions to be isotonic only requirement is that their osmotic pressures should be same. There are no conditions on solutes and solvents.

Question 14. If

α

is the degree of dissociation of

N a_{2} S O_{4}

, the vant Hoff's factor (i) used for calculating the molecular mass is

1+α
α
1+2α
1−2α

Discuss Question

Answer: Option C. -> 1+2α
:
C

N a_{2} S O_{4}      1 - α ⇌ 2 N a^{+}      2 α + S O_{4}^{2 -}      α

Totalno. ofionsafter dissociation =

1 + 2 α

vant Hoff's factor,

i = \frac{N o . o f p a r t i c l e s a f t e r d i s s o c i a t i o n}{N o . o f p a r t i c l e s b e f o r e d i s s o c i a t i o n} = 1 + 2 α

Question 15. Which of the followings is a colligative property

Osmotic pressure
Boiling point
Vapour pressure
Freezing point

Discuss Question

Answer: Option A. -> Osmotic pressure
:
A
Only Osmotic pressure is a colligative property.
‘Relative lowering of vapour pressure of the solvent’ is a colligative property but not only vapour pressure. Similar argument for boiling point and freezing point

Question 16. Osmotic pressure of a solution containing 0.1 mole of solute per litre at 273K is (in atm)

0.11×0.08205×273
0.1×1×0.08205×273
10.1×0.08205×273
0.11× 2730.08205

Discuss Question

Answer: Option A. -> 0.11×0.08205×273
:
A
Osmoticpressure,

π = C R T = \frac{n}{V} R T = \frac{0.1}{1} \times 0.08205 \times 273 a t m

Question 17. The average osmotic pressure of human blood is 7.8 bar at 37⁰C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream

0.16 mol/L
0.32mol/L
0.60mol/L
0.45mol/L

Discuss Question

Answer: Option B. -> 0.32mol/L
:
B

Question 18. A 0.001 molal solution of

[P t (N H_{3})_{4} c l_{4}]

in water had a freezing point depression of

{0.0054}^{\circ}

. If

K_{f}

for water is 1.80, the correct formulation for the above molecule is

[Pt(NH3)4Cl3]Cl
[Pt(NH3)4]Cl4
[Pt(NH3)4Cl2]Cl3
[Pt(NH3)4Cl2]Cl2

Discuss Question

Answer: Option D. -> [Pt(NH3)4Cl2]Cl2
:
D

△ T = i K_{f} m

\Rightarrow i = \frac{△ T_{f}}{K_{f} m} = \frac{0.0054}{1.80 \times 0.001} = 3

Thus complexproducesthreeionspermolecule.hence possiblecorrectformulais

[P t (N H_{3})_{4} C l_{2}] C l_{2}

Question 19. The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be