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Mass percentage of CCl
4
Alternatively,
Mass percentage of CCl
4
= (100 − 15.28)%
= 84.72%
= 0.3846 mol
Molar mass of carbon tetrachloride (CCl
4
) = 1 × 12 + 4 × 35.5
= 154 g mol
−1
∴Number of moles of CCl
4
= 70 / 154
= 0.4545 mol
Thus, the mole fraction of C
6
H
6
is given as:
= 0.458
(a) Molar mass of Co (NO
3
)
2
.6H
2
O = 59 + 2 (14 + 3 × 16) + 6 × 18
= 291 g mol
−1
∴Moles of Co (NO
3
)
2
.6H
2
O
= 0.103 mol
Therefore, molarity
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H
2
SO
4
= 0.5 mol
∴Number of moles present in 30 mL of 0.5 M H
2
SO
4
= 0.015 mol
Therefore, molarity
= 0.03 M
= 36.95 g
= 37 g of urea (approximately)
Hence, mass of urea required = 37 g
Question
5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL
-1.
Solution :
(a) Molar mass of KI = 39 + 127 = 166 g mol
−1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 − 20) g of water = 80 g of water
Therefore, molality of the solution
= 1.506 m
= 1.51 m (approximately)
(b) It is given that the density of the solution = 1.202 g mL
−1
∴Volume of 100 g solution
= 83.19 mL
= 83.19 × 10
−3 L
Therefore, molarity of the solution
= 1.45 M
(c) Moles of KI
Moles of water
Therefore, mole fraction of KI
= 0.0263
Question
6. H
2
S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H
2
S in water at STP is 0.195 m, calculate Henry’s law constant.
Solution :
It is given that the solubility of H
2
S in water at STP is 0.195 m, i.e., 0.195 mol of H
2
S is dissolved in 1000 g of water.
Moles of water
= 55.56 mol
∴Mole fraction of H
2
S, x
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law:
p = K
H
x
= 282 bar
Question
7. Henry’s law constant for CO
2
in water is 1.67 × 10
8
Pa at 298 K. Calculate the quantity of CO
2
in 500 mL of soda water when packed under 2.5 atm CO
2
pressure at 298 K.
Solution :
It is given that:
K
H
= 1.67 × 10
8 Pa
= 2.5 atm = 2.5 × 1.01325 × 10
5 Pa
= 2.533125 × 10
5 Pa
According to Henry’s law:
= 0.00152
We can write,
[Since,
is negligible as compared to
]
In 500 mL of soda water, the volume of water = 500 mL
[Neglecting the amount of soda present]
We can write:
500 mL of water = 500 g of water
nH
2
0= 27.78 mol of water
Now,
Hence, quantity of CO
2
in 500 mL of soda water = (0.042 × 44)g
= 1.848 g
Question
8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Solution :
It is given that:
= 450 mm of Hg
= 700 mm of Hg
p
total
= 600 mm of Hg
From Raoult’s law, we have:
Therefore, total pressure,
Therefore,
= 1 − 0.4
= 0.6
Now,
= 450 × 0.4
= 180 mm of Hg
= 700 × 0.6
= 420 mm of Hg
Now, in the vapour phase:
Mole fraction of liquid A
= 0.30
And, mole fraction of liquid B = 1 − 0.30
= 0.70
Question
9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH
2
CONH
2
) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution :
It is given that vapour pressure of water,
= 23.8 mm of Hg
Weight of water taken, w
1
= 850 g
Weight of urea taken, w
2
= 50 g
Molecular weight of water, M
1
= 18 g mol
−1
Molecular weight of urea, M
2
= 60 g mol
−1
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p
1
.
Now, from Raoult’s law, we have:
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.
Question
10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol
−1.
Solution :
Here, elevation of boiling point ΔT
b
= (100 + 273) − (99.63 + 273)
= 0.37 K
Mass of water, w
l
= 500 g
Molar mass of sucrose (C
12
H
22
O
11
), M
2
= 11 × 12 + 22 × 1 + 11 × 16
= 342 g mol
−1
Molal elevation constant, K
b
= 0.52 K kg mol
−1
We know that:
= 121.67 g (approximately)
Hence, 121.67 g of sucrose is to be added.
Question
11. Calculate the mass of ascorbic acid (Vitamin C, C
6
H
8
O
6
) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K
f
= 3.9 K kg mol
−1.
Solution :
Mass of acetic acid, w
1
= 75 g
Molar mass of ascorbic acid (C
6
H
8
O
6
), M
2
= 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol
−1
Lowering of melting point, ΔT
f
= 1.5 K
We know that:
= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid is needed to be dissolved.
Question
12. 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 37°C.
Solution :
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)
Question
13. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Solution :
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.
Question
14. Give an example of solid solution in which the solute is a gas.
Solution :
In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.
Question
15. Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Solution :
(i) Mole fraction:
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.
i.e.,
Mole fraction of a component
Mole fraction is denoted by ‘x’.
If in a binary solution, the number of moles of the solute and the solvent are n
A
and n
B
respectively, then the mole fraction of the solute in the solution is given by,
Similarly, the mole fraction of the solvent in the solution is given as:
(ii) Molality
Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:
Molality (m)
(iii) Molarity
Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution.
It is expressed as:
Molarity (M)
(iv) Mass percentage:
The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution. It is expressed as:
Mass % of a component
Question
15. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL
−1?
Solution :
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO
3
) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol
−1
Then, number of moles of HNO
3
= 1.079 mol
Given,
Density of solution = 1.504 g mL
−1
(Mass / Density) = Volume of 100 g solution =
Molarity of solution
Question
16. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL
−1,
then what shall be the molarity of the solution?
Solution :
10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 − 10) g = 90 g of water.
Molar mass of glucose (C
6
H
12
O
6
) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol
−1
Then, number of moles of glucose
= 0.056 mol
⇒ Molality of solution
= 0.62 m
Number of moles of water
= 5 mol
⇒ Mole fraction of glucose
And, mole fraction of water
= 1 − 0.011
= 0.989
If the density of the solution is 1.2 g mL
−1
,
then the volume of the 100 g solution can be given as:
⇒ Molarity of the solution
= 0.67 M
Question
17. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na
2
CO
3
and NaHCO
3
containing equimolar amounts of both?
Solution :
Let the amount of Na
2
CO
3
in the mixture be x g.
Then, the amount of NaHCO
3
in the mixture is (1 − x) g.
Molar mass of Na
2
CO
3
= 2 × 23 + 1 × 12 + 3 × 16
= 106 g mol
−1
⇒ Number of moles Na
2
CO
3
Molar mass of NaHCO
3
= 1 × 23 + 1 × 1 × 12 + 3 × 16
= 84 g mol
−1
⇒ Number of moles of NaHCO
3
According to the question,
⇒ 84x = 106 − 106x
⇒ 190x = 106
⇒ x = 0.5579
Therefore, number of moles of Na
2
CO
3
= 0.0053 mol
And, number of moles of NaHCO
3
= 0.0053 mol
HCl reacts with Na
2
CO
3
and NaHCO
3
according to the following equation.
1 mol of Na
2
CO
3
reacts with 2 mol of HCl.
Therefore, 0.0053 mol of Na
2
CO
3
reacts with 2 × 0.0053 mol = 0.0106 mol.
Similarly, 1 mol of NaHCO
3
reacts with 1 mol of HCl.
Therefore, 0.0053 mol of NaHCO
3
reacts with 0.0053 mol of HCl.
Total moles of HCl required = (0.0106 + 0.0053) mol
= 0.0159 mol
In 0.1 M of HCl,
0.1 mol of HCl is preset in 1000 mL of the solution.
Therefore, 0.0159 mol of HCl is present in
= 159 mL of the solution
Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na
2
CO
3
and NaHCO
3,
containing equimolar amounts of both.
Question
18. 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.
Solution :
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%
Question
19. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C
2
H
6
O
2
) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL
−1
, then what shall be the molarity of the solution?
Solution :
Molar mass of ethylene glycol
= 2 × 12 + 6 × 1 + 2 ×16
= 62 gmol
−1
Number of moles of ethylene glycol
= 3.59 mol
Therefore, molality of the solution
= 17.95 m
Total mass of the solution = (222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 g mL
−1
⇒ Volume of the solution
= 394.22 mL
= 0.3942 × 10
−3 L
∴ Molarity of the solution
= 9.11 M
Question
20. A sample of drinking water was found to be severely contaminated with chloroform (CHCl
3
) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Solution :
(i) 15 ppm (by mass) means 15 parts per million (10
6) of the solution.
Therefore, percent by mass
= 1.5 × 10
−3 %
(ii) Molar mass of chloroform (CHCl
3
) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5 g mol
−1
Now, according to the question,
15 g of chloroform is present in 10
6
g of the solution.
i.e., 15 g of chloroform is present in (10
6
− 15) ≈
106
g of water.
∴ Molality of the solution
= 1.26 × 10
−4 m
Question
21. What role does the molecular interaction play in a solution of alcohol and water?
Solution :
In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.
Question
22. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Solution :
Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.
Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.
Question
23. State Henry’s law and mention some important applications?
Solution :
Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as:
p = K
H
x
Where,
K
H
is Henry’s law constant
Some important applications of Henry’s law are mentioned below.
(i) Bottles are sealed under high pressure to increase the solubility of CO
2
in soft drinks and soda water.
(ii) Henry’s law states that the solubility of gases increases with an increase in pressure. Therefore, when a scuba diver dives deep into the sea, the increased sea pressure causes the nitrogen present in air to dissolve in his blood in great amounts. As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading to the formation of nitrogen bubbles in the blood. This results in the blockage of capillaries and leads to a medical condition known as ‘bends’ or ‘decompression sickness’.
Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.
(iii) The concentration of oxygen is low in the blood and tissues of people living at high altitudes such as climbers. This is because at high altitudes, partial pressure of oxygen is less than that at ground level. Low-blood oxygen causes climbers to become weak and disables them from thinking clearly. These are symptoms of anoxia.
Question
24. The partial pressure of ethane over a solution containing 6.56 × 10
−3
g of ethane is 1 bar. If the solution contains 5.00 × 10
−2
g of ethane, then what shall be the partial pressure of the gas?
Solution :
Molar mass of ethane (C
2
H
6
) = 2 × 12 + 6 × 1
= 30 g mol
−1
∴Number of moles present in 6.56 × 10
−3 g of ethane
= 2.187 × 10
−4 mol
Let the number of moles of the solvent be x.
According to Henry’s law,
p = K
H
x
Number of moles present in 5.00 × 10
−2
g of ethane
= 1.67 × 10
−3
mol
According to Henry’s law,
p = K
H
x
Vapour pressure of a two-component solution showing positive deviation from Raoult’s law
Vapour pressure of a two-component solution showing negative deviation from Raoult’s law
In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.
Δ
sol
H = 0
In the case of solutions showing positive deviations, absorption of heat takes place.
∴Δ
sol
H = Positive
In the case of solutions showing negative deviations, evolution of heat takes place.
∴Δ
sol
H = Negative
Question
26. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Solution :
Here,
Vapour pressure of the solution at normal boiling point (p
1
) = 1.004 bar
Vapour pressure of pure water at normal boiling point
Mass of solute, (w
2
) = 2 g
Mass of solvent (water), (w
1
) = 98 g
Molar mass of solvent (water), (M
1
) = 18 g mol
−1
According to Raoult’s law,
= 41.35 g mol
−1
Hence, the molar mass of the solute is 41.35 g mol
−1.
Question
27. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Solution :
Vapour pressure of heptane
Vapour pressure of octane
= 46.8 kPa
We know that,
Molar mass of heptane (C
7
H
16
) = 7 × 12 + 16 × 1
= 100 g mol
−1
∴ Number of moles of heptane
= 0.26 mol
Molar mass of octane (C
8
H
18
) = 8 × 12 + 18 × 1
= 114 g mol
−1
∴ Number of moles of octane = 35 /114 mol
= 0.31 mol
Mole fraction of heptane,
= 0.456
And, mole fraction of octane, x
2
= 1 − 0.456
= 0.544
Now, partial pressure of heptane,
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane,
= 0.544 × 46.8
= 25.46 kPa
Hence, vapour pressure of solution, p
total
= p
1
+ p
2
= 47.97 + 25.46
= 73.43 kPa
Question
28. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Solution :
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol
−1
= Number of moles present in 1000 g of water
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
.
It is given that,
Vapour pressure of water,
= 12.3 kPa
Applying the relation,
⇒ 12.3 − p
1
= 0.2177
⇒ p
1
= 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
Question
29. Calculate the mass of a non-volatile solute (molar mass 40 g mol
−1
) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Solution :
Let the vapour pressure of pure octane be
Molar mass of solute, M
2
= 40 g mol
−1
Mass of octane, w
1
= 114 g
Molar mass of octane, (C
8
H
18
), M
1
= 8 × 12 + 18 × 1
= 114 g mol
−1
Applying the relation,
Hence, the required mass of the solute is 8 g.
Question
30. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
molar mass of the solute
vapour pressure of water at 298 K.
Solution :
(i) Let, the molar mass of the solute be M g mol
−1
Now, the no. of moles of solvent (water),
And, the no. of moles of solute,
Applying the relation:
After the addition of 18 g of water:
Again, applying the relation:
Dividing equation (i) by (ii), we have:
Therefore, the molar mass of the solute is 23 g mol
−1.
(ii) Putting the value of ‘M’ in equation (i), we have:
.gif)
Hence, the vapour pressure of water at 298 K is 3.53 kPa.
Page No 60:
Question
31. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Solution :
Here, ΔT
f
= (273.15 − 271) K
= 2.15 K
Molar mass of sugar (C
12
H
22
O
11
) = 12 × 12 + 22 × 1 + 11 × 16
= 342 g mol
−1
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.
Now, number of moles of cane sugar
= 0.0146 mol
Therefore, molality of the solution,
= 0.1537 mol kg
−1
Applying the relation,
ΔT
f
= K
f
× m
= 13.99 K kg mol
−1
Molar of glucose (C
6
H
12
O
6
) = 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol
−1
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.
∴ Number of moles of glucose
= 0.0278 mol
Therefore, molality of the solution,
= 0.2926 mol kg
−1
Applying the relation,
ΔT
f
= K
f
× m
= 13.99 K kg mol
−1 × 0.2926 mol kg−1
= 4.09 K (approximately)
Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K.
Question
32. Two elements A and B form compounds having formula AB
2
and AB
4
. When dissolved in 20 g of benzene (C
6
H
6
), 1 g of AB
2
lowers the freezing point by 2.3 Kwhereas 1.0 g of AB
4
lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol
−1
. Calculate atomic masses of A and B.
Solution :
We know that,
Then,
= 110.87 g mol
−1
= 196.15 g mol
−1
Now, we have the molar masses of AB
2
and AB
4
as 110.87 g mol
−1 and 196.15 g mol−1 respectively.
Let the atomic masses of A and B be x and y respectively.
Now, we can write:
Subtracting equation (i) from (ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of ‘y’ in equation (1), we have
x + 2 × 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.
Question
33. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Solution :
Here,
T = 300 K
π = 1.52 bar
R = 0.083 bar L K
−1 mol−1
Applying the relation,
π = CRT
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
Question
34. Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) I
2
and CCl
4
(iii) NaClO
4
and water
(iv) methanol and acetone
(v) acetonitrile (CH
3
CN) and acetone (C
3
H
6
O).
Solution :
(i) Van der Wall’s forces of attraction.
(ii) Van der Wall’s forces of attraction.
(iii) Ion-diople interaction.
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.
Question
35. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH
3
OH, CH
3
CN.
Solution :
n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.
The order of increasing polarity is:
Cyclohexane < CH
3
CN < CH
3
OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH
3
OH < CH
3
CN < Cyclohexane
Question
36. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Solution :
(i) Phenol (C
6
H
5
OH) has the polar group −OH and non-polar group −C
6
H
5
. Thus, phenol is partially soluble in water.
(ii) Toluene (C
6
H
5
−CH
3
) has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water. Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol
has polar −OH group and can form H−bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol (C
5
H
11
OH) has polar −OH group, but it also contains a very bulky non-polar −C
5
H
11
group. Thus, pentanol is partially soluble in water.
Question
37. If the density of some lake water is 1.25 g mL
−1
and contains 92 g of Na
+
ions per kg of water, calculate the molality of Na
+
ions in the lake.
Solution :
Number of moles present in 92 g of Na
+
ions =
= 4 mol
Therefore, molality of Na
+ ions in the lake
= 4 m
Question
38. If the solubility product of CuS is 6 × 10
−16
, calculate the maximum molarity of CuS in aqueous solution.
Solution :
Solubility product of CuS, K
sp
= 6 × 10
−16
Let s be the solubility of CuS in mol L
−1.
Now,
= s × s
= s
2
Then, we have, K
sp
=
= 2.45 × 10
−8 mol L−1
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10
−8
mol L
−1.
Question
39. Calculate the mass percentage of aspirin (C
9
H
8
O
4
) in acetonitrile (CH
3
CN) when 6.5 g of C
9
H
8
O
4
is dissolved in 450 g of CH
3
CN.
Solution :
6.5 g of C
9
H
8
O
4
is dissolved in 450 g of CH
3
CN.
Then, total mass of the solution = (6.5 + 450) g
= 456.5 g
Therefore, mass percentage ofC
9
H
8
O
4
= 1.424%
Question
40. Nalorphene (C
19
H
21
NO
3
), 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 1.5 × 10
−3
m aqueous solution required for the above dose.
Solution :
The molar mass of nalorphene
is given as:
In 1.5 × 10
−3
m aqueous solution of nalorphene,
1 kg (1000 g) of water contains 1.5 × 10
−3 mol
= 0.4665 g
Therefore, total mass of the solution
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:
Hence, the mass of aqueous solution required is 3.22 g.
Question
41. Calculate the amount of benzoic acid (C
6
H
5
COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution :
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 (C
6
H
5
COOH) = 7 × 12 + 6 × 1 + 2 × 16
= 122 g mol
−1
Hence, required benzoic acid = 0.0375 mol × 122 g mol
−1
= 4.575 g
Question
42. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Solution :
Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H
+
ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid
Question
43. Calculate the depression in the freezing point of water when 10 g of CH
3
CH
2
CHClCOOH is added to 250 g of water. K
a
= 1.4 × 10
−3,
K
f
= 1.86 K kg mo
l−1.
Solution :
Molar mass of
∴No. of moles present in 10 g of
= 0.0816 mol
It is given that 10 g of
is added to 250 g of water.
∴Molality of the solution,
Let α be the degree of dissociation of
undergoes dissociation according to the following equation:
Since α is very small with respect to 1, 1 − α ≈ 1
Now,
Again,
Total moles of equilibrium = 1 − α + α + α
= 1 + α
Hence, the depression in the freezing point of water is given as:
Question
44. 19.5 g of CH
2
FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Solution :
It is given that:
We know that:
Therefore, observed molar mass of
The calculated molar mass of
is:
Therefore, van’t Hoff factor,
Let α be the degree of dissociation of
Now, the value of K
a
is given as:
Taking the volume of the solution as 500 mL, we have the concentration:
Therefore,
Question
45. Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.
Solution :
Vapour pressure of water,
= 0.139 mol
And, number of moles of water,
= 25 mol
We know that,
⇒ 17.535 − p
1
= 0.097
⇒ p
1
= 17.44 mm of Hg
Hence, the vapour pressure of water is 17.44 mm of Hg.
Question
46. Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 10
5
mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.
Solution :
Here,
p = 760 mm Hg
k
H
= 4.27 × 10
5 mm Hg
According to Henry’s law,
p = k
H
x
= 177.99 × 10
−5
= 178 × 10
−5 (approximately)
Hence, the mole fraction of methane in benzene is 178 × 10
−5.
Question
47. 100 g of liquid A (molar mass 140 g mol
−1
) was dissolved in 1000
g of liquid B (molar mass 180 g mol
−1
). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Solution :
Number of moles of liquid A,
= 0.714 mol
Number of moles of liquid B,
= 5.556 mol
Then, mole fraction of A,
= 0.114
And, mole fraction of B, x
B
= 1 − 0.114
= 0.886
Vapour pressure of pure liquid B,
= 500 torr
Therefore, vapour pressure of liquid B in the solution,
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution, p
total
= 475 torr
∵ Vapour pressure of liquid A in the solution,
p
A
= p
total
− p
B
= 475 − 443
= 32 torr
Now,
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
Question
48. Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p
total
’ p
chloroform
’ and p
acetone
as a function of x
acetone
. The experimental data observed for different compositions of mixture is.
It can be observed from the graph that the plot for the p
total
of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.
Question
49. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Solution :
Molar mass of benzene
Molar mass of toluene
Now, no. of moles present in 80 g of benzene
= 1.026
And, no. of moles present in 100 g of toluene = 100/92 = 1.087 mol
∴Mole fraction of benzene, x
b
And, mole fraction of toluene,
It is given that vapour pressure of pure benzene,
And, vapour pressure of pure toluene,
Therefore, partial vapour pressure of benzene,
And, partial vapour pressure of toluene,
Hence, mole fraction of benzene in vapour phase is given by:
Question
50. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 10
7
mm and 6.51 × 10
7
mm respectively, calculate the composition of these gases in water.
Solution :
Percentage of oxygen (O
2
) in air = 20 %
Percentage of nitrogen (N
2
) in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg = 7600 mm Hg
Therefore,
Partial pressure of oxygen,
= 1520 mm Hg
Partial pressure of nitrogen,
= 6004 mmHg
Now, according to Henry’s law:
p = K
H
.x
For oxygen:
For nitrogen:
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10
−5and 9.22 × 10−5 respectively.
Question
51. Determine the amount of CaCl
2
(i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Solution :
We know that,
Here,
R = 0.0821 L atm K
-1mol-1
M = 1 × 40 + 2 × 35.5
= 111g mol
-1
Therefore, w
= 3.42 g
Hence, the required amount of CaCl
2
is 3.42 g.
Question
52. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K
2
SO
4
in 2 liter of water at 25° C, assuming that it is completely dissociated.
Solution :
When K
2
SO
4
is dissolved in water,
ions are produced.
⇒ Total number of ions produced = 3
⇒ i =3
Given,
w = 25 mg = 0.025 g
V = 2 L
T = 25
0C = (25 + 273) K = 298 K
Also, we know that:
R = 0.0821 L atm K
-1mol-1
M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol
-1
Appling the following relation,
