CBSE Class 12 Chemistry Chapter 2 Important Questions – Free PDF Download
Free PDF download of Important Questions for CBSE Class 12 Chemistry Chapter 2 – Solutions 3 Marks Questions prepared by expert Chemistry teachers from latest edition of CBSE(NCERT) books, On CoolGyan.Org to score more marks in CBSE board examination.
You can also Download Chemistry Revision Notes Class 12 to help you to revise complete Syllabus and score more marks in your examinations.
CBSE Class 12 Chemistry Important Questions Chapter 2 – Solutions
3 Marks Questions
1. Find the molality and molarity of a 15% solution of 
when its density is 1.10 
& molar mass = 98 amu.
Ans. Volume = mass/density
Molarity = 
= 
.
Molality = 
= 1.8 M.
2. Calculate the mole fraction of ethanol and water in a sample of rectified spirit which contains 46% ethanol by mass?
Ans. Mass of ethanol = 46g
Mass of water = 100 – 46 = 54g
Mole fraction of ethanol, 
= 
Mole fraction of water = 1-0.25 = 0.75
3. Calculate the % composition in terms of mass of a solution obtained by mixing 300g of a 25% & 400 g of a 40% solution by mass?
Ans. mass of solute in 400g of 40% = 
Total mass of solute = 160+75 = 235g
Total mass of solution = 400+300 = 700g
Mass% of solute = 
= 
Mass % of solvent = 100 – 33.57 = 66.43%
4. One litre of sea water weight 1030g and contains about 
of dissolved
. Calculate the concentration of dissolved oxygen in ppm?
Ans. mass of 
ppm of 
in 1030 g sea water = 
= 
5. The density of 85% phosphoric acid is
. What is the volume of a solution that contains 17g of phosphoric acid?
Ans. 85g phosphoric acid is present in 100g of solution.
17g of phosphoric acid is present in 
Volume of 17g of 85% acid = 
= 
6. Define the term azeotrope?
Ans. A solution at certain concentration when continues to boil at constant temperature without change in its composition in solution & in vapour phase is called an azeotrope.
7. Obtain a relationship between relative lowering of vapour pressure and mole fraction of solute?
Ans. According to Raoult’s Law –
= 
Relative lowering of vapour pressure.
8. Draw the graphs of both deviations from ideal behaviours?
Ans.
9. A weak electrolyte AB in 5% dissociated in aqueous solution? What is the freezing point of a 0.10 molar aqueous solution of AB? Kf = 1.86 deg/molal?
Ans. Degree of dissociation 
of AB = 
.
AB 
A+ + B–
M 0 0
No. of moles dissolved
No. of moles after dissociations m(1-
) m m
0.1 (1 – 0.05) 
Total moles = [0.1(1- 0.05)] + 
+ 
= 0.095 + 0.005 + 0.005 = 0.105m
= 0.1953 deg.
10. Henry’s law constant for the molality of methane in benzene at 298 K is 
. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Ans. Here,
p = 760 mm Hg
According to Henry’s law,
= 
= 
(approximately)
Hence, the mole fraction of methane in benzene is .
11. Nalorphene 
, similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 
m aqueous solution required for the above dose.
Ans. The molar mass of nalorphene is given as:
In 
aqueous solution of nalorphene,
1 kg (1000 g) of water contains mol = 
= 0.4665 g
Therefore, total mass of the solution = (1000 + 0.4665) g
= 1000.4665 g
This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.
Therefore, mass of the solution containing 1.5 mg of nalorphene is:
= 3.22 g
Hence, the mass of aqueous solution required is 3.22 g.
12. Calculate the amount of benzoic acid 
required for preparing 250 mL of 0.15 M solution in methanol.
Ans. 0.15 M solution of benzoic acid in methanol means,
1000 mL of solution contains 0.15 mol of benzoic acid
Therefore, 250 mL of solution contains = 
mol of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid 
= 122 g 
Hence, required benzoic acid = 
= 4.575 g
13. If the solubility product of CuS is
, calculate the maximum molarity of CuS in aqueous solution.
Ans. Solubility product of CuS, 
Let s be the solubility of CuS in
.
Now, 
= 
Then, we have, 
= 2.45 × 10 – 8
Hence, the maximum molarity of CuS in an aqueous solution is
.
14. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Ans. Total amount of solute present in the mixture is given by,
= 75 + 160
= 235 g
Total amount of solution = 300 + 400 = 700 g
Therefore, mass percentage (w/w) of the solute in the resulting solution, 
= 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution,
= (100 – 33.57)%
= 66.43%
15. Calculate the mass of ascorbic acid (Vitamin C,
) to be dissolved in 75 g of acetic acid to lower its melting point by
.
.
Ans. Mass of acetic acid, 
Molar mass of ascorbic acid 
= 
Lowering of melting point, 
We know that:
= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid is needed to be dissolved.
16. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at
.
Ans. It is given that:
Volume of water, V= 450 mL = 0.45 L
Temperature, T = (37 + 273)K = 310 K
Number of moles of the polymer, 
We know that:
Osmotic pressure, 
= 30.98 Pa
= 31 Pa (approximately)
17. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Ans. Homogeneous mixtures of two or more than two components are known as solutions.
There are three types of solutions.
(i) Gaseous solution:
The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.
(ii) Liquid solution:
The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid.
For example, a solution of ethanol in water is a liquid solution.
(iii) Solid solution:
The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.
18. Calculate the mass of urea (
) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans. Molar mass of urea 
= 60 
0.25 molar aqueous solution of urea means:
1000 g of water contains 0.25 mol = 
of urea
= 15 g of urea
That is,
(1000 + 15) g of solution contains 15 g of urea
Therefore, 2.5 kg (2500 g) of solution contains =
= 36.95 g
= 37 g of urea (approximately)
Hence, mass of urea required = 37 g