CBSE Class 12 Physics Chapter 2 – Solutions Important Question Answer
Very Short Type Question – 1 Mark
1. Define the term –solubility?
Ans. The maximum amount of a substance that can be dissolved in a specified amount of solvent is called its solubility.
2. What is the effect of pressure on solubility of a gas?
Ans. The solubility of a gas increases with increases of pressure.
3. State Henry’s Law.
Ans. Henry’s Law states that at a constant temperature the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
4. State Raoult’s Law.
Ans. Raoult’s Law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
5. What are the factors on which vapour pressure depends?
Ans. The factors on which vapour pressure depends are –
1) Temperature of the liquid. 2) Nature of the liquid.
6. The vapour pressure of solvent gets lowered, when a non- volatile solute is
added to it. Why?
Ans. When a non-volatile solute is added to a solvent, the surface area for escape of solvent molecules decreases and vapour pressure gets lowered.
7. Name two ways by which vapour pressure of a liquid can be lowered.
Ans. The two ways by which vapour pressure can be lowered are –
1) By decreasing the temperature.
2) By adding a non- volatile solute.
8. Define‘solution’?
Ans. Solutions are homogeneous mixtures of two or more than two components.
9. Define the following terms : (a) Molality (b) Molarity
Ans. (a) Molality is defined as the number of moles of the solute per kilogram of solvent.
(b) Molarity (M) = Number of moles of solute dissolved in one litre of solution.
Molarity (M)
10. How does change in temperature changes the molarity and molality values?
Ans. As the temperature increases, volume increases and molarity decreases whereas molality does not change with any change in temperature.
11. Define the term colligative properties?
Ans. The properties which depends upon amount of solute and not upon the nature of solute are called colligative properties.
12. What are the possible deviations from ideal behaviors?
Ans. There are two types of deviation from ideal behaviour – positive and negative deviations.
13. Give one example of each deviation?
Ans. Positive deviation – ethanol and acetone.
Negative deviation – chloroform and acetone.
14. At 
, the vapour pressure of water is 55.3 mm Hg .Calculate the vapour pressure at the same temperature over 10% aqueous solution of urea
?
Ans. 53.48 mm Hg.
15. How much urea (molar mass 60 g/mol) should be dissolved in 50g of water so that its vapour pressure at room temperature is reduced by 25%?
Ans. 41.7 g .
16. Why is the boiling point elevated when a non – volatile solute is dissolved in a liquid?
Ans. When a non – volatile solute is added the vapour pressure decreases and the solution is heated to a higher temperature, increasing the boiling point.
17. How is boiling point changed when mass of solvent is doubled?
Ans. 
= when the amount of solvent is doubled, ∆Tb is halved.
18. What happens when red blood cells are placed in 0.1% NaCl solution?
Ans. Water from NaCl solution passes into cells &they swell. Finally they will burst.
19. How is osmotic pressure of a solution related to its concentration?
Ans. Osmotic pressure, π = CRT
C = concentration ,R = gas constant.
T= temperature
20. Calculate the osmotic pressure of 0.25 M solution of urea at
. R = 0.083 L bar/mol/k.
Ans. T = = 310k
= 
RT
= 6.43 bar.
21. An aqueous solution of glucose, 
has osmotic pressure of 2.72 atm at 298k. How many moles of glucose were dissolved per litre of solution?
Ans. 0.111 mol.
22. When does the measurement of colligative property leads to abnormal molecular mass?
Ans. When the solute undergoes either association or disassociation abnormal molar mass is obtained.
23. When is the value of i less then unity?
Ans. When the solute under goes association in solution, I is less then unity.
24. The molecular mass of a solute is 120 g/mol and van’t Hoff factor is 4. What is its abnormal molecular mass?
Ans. Abnormal molecular mass = 
= 
.
Short type question answer – 2 Marks
1. Calculate the volume of water which could be added to 20 ml of 0.65 m HCl to dilute the solution to 0.2 m?
Ans.For dilution –
Vol of water to be added to 20 ml = 
= 65ml – 20ml = 45 ml.
2. A solution is prepared by dissolving 11g glucose in 
water at
. What is the mass Percentage of glucose in solution? The density of water is
?
Ans.Density = 
0.996 = 
= 199.2 g
Mass% of glucose = 
= 
3. Carbon tetrachloride and water are immiscible whereas alcohol and water are miscible. Explain on the basis of molecular structures of there compounds.
Ans.Carbon tetrachloride is a non-polar compound whereas water is a polar compound. They do not interact with each other and carbon tetrachloride cannot dissolve in water whereas alcohol and water are completely miscible due to high polarity.
4. Why do mountaineers carry oxygen cylinder while climbing mountains?
Ans.At high attitudes the partial pressure of oxygen is less than that of the ground level which decreases the concentration of oxygen in blood and tissues. Low blood oxygen causes climbers to become weak and unable to think clearly & they suffer from anoxia. To avoid such situations, mountaineers carry oxygen cylinder while climbing.
5. Plot a graph between vapour pressure and mole fraction of a solution obeying Raoult’s Law at constant temperature?
Ans.
;
6. Name different colligative properties?
Ans. The colligative properties are –
a) Relative lowering of vapour pressure.
b) Elevation in boiling point.
c) Depression in freezing point
d) Osmotic pressure.
7. Give the characteristics of ideal solution?
Ans. An ideal solution is formed from two liquids only when –
a) They obey Raoult’s Law
b) 
Hmix = 0
c) Vmix = 0
d) The various inter molecular forces are identical.
8. A mixture of chlorobenzene and bromobenzene is a nearly an ideal solution but a mixture of chloroform and acetone is not Explain?
Ans. Chlorobezene & bromobenzene both have similar structure and polarity. Therefore the various interactions (solute – solute, solvent – solvent & solute – solvent) are same whereas in chloroform and acetone initially there is no hydrogen bonding but after mixing solute solvent interactions (H –bond ) become stronger and solution deviates from ideal behaviour.
9. 0.90g of a non – electrolyte was dissolved in 87.90g of benzene. This raised the boiling point of benzene by
. If the molecular mass of non – electrolyte is 103.0 g/mol, calculate the molal elevation constant for benzene?
Ans. 
= 2.514 k kg/mol.
10. Show graphically the depression in freezing point on adding a non volatile solute?
Ans.
11. Define cryoscopic constant?
Ans. When 1 mole of a solute (that neither dissociates nor associates) is dissolved in 1kg of solvent, the depression in freezing point is called cryoscopic constant.
12. When 20g of a non – volatile solid is added to 250 ml of water, the freezing point of water becomes
. Calculate molecular mass of the solid if kf of water is
.
Ans. 250 ml = 250g as density of water = 1g/ml
Freezing pt of water = 273k
Freezing point of solution = – 0.90C+273 = 272.1k
ΔTf = 273 k – 272.1k = 0.9k
Msolute = 
= 
.
13. Give various expressions for van’t Hoff factor?
Ans.i=
=
=
14. How are the various colligative properties modified after consideration of van’t Hoff factor?
Ans. a) 
b) 
c) 
d) 
15. The boiling point elevation of 0.6 g acetic acid in 100g benzene is 0.1265k. What conclusion can you draw about the state of solute in solution? Molar elevation constant for benzene is 2.53 deg per molar?
Ans. Molality of acetic acid = 
= 0.10 m.
M = 
i = 
Since i =
, acetic acid exist an dimer in solution.
16. The osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm at
. Calculate van’t Hoff factor. R=0.082 L atm/1 mol/k?
Ans. i = 2.76 The solute is dissociated in solution.
17. Calculate the mass percentage of aspirin 
in acetonitrile 
when 6.5 g of 
is dissolved in 450 g of 
.
Ans. 6.5 g of is dissolved in 450 g of .
Then, total mass of the solution = (6.5 + 450) g
= 456.5 g
Therefore, mass percentage of = 
= 1.424%
18. If the density of some lake water is 
and contains 92 g of 
ions per kg of water, calculate the molality of ions in the lake.
Ans. Number of moles present in 92 g of ions = 
= 4 mol
Therefore, molality of Na+ ions in the lake 
= 4 m
19. 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?
Ans. Here,
T = 300 K
= 1.52 bar
R = 0.083 bar L 
Applying the relation,
= CRT
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
20. Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) 
and 
(iii) 
and water
(iv) methanol and acetone
(v) acetonitrile 
and acetone 
.
Ans. (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.
21. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl,
,.
Ans. 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 < < < KCl
Therefore, the order of increasing solubility is:
KCl < < < Cyclohexane
22. A sample of drinking water was found to be severely contaminated with chloroform (
) 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.
Ans. (i) 15 ppm (by mass) means 15 parts per million (106) of the solution.
Therefore, percent by mass =
= 
(ii) Molar mass of chloroform 
= 
Now, according to the question,
15 g of chloroform is present in 106 g of the solution.
i.e., 15 g of chloroform is present in 
106 g of water.
Therefore, Molality of the solution = 
= 
23. What role does the molecular interaction play in a solution of alcohol and water?
Ans. 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.
24. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Ans. 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.
25. Give an example of solid solution in which the solute is a gas.
Ans. 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.
Short type question answer – 3 Marks
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. 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.
15. 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)
16. 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.
17. 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
Long term question answer – 5 Marks
1. The vapour pressure of 
at 
is 854 mm Hg .A solution of 2.0g sulphur in 100g of has a vapour pressure of 848.9 mm Hg .Calculate the formula
of sulphur molecule.
Ans. 
= 854 mm
, 
,
= 
= 
= 254.5 g/mol.
Let the formula = Sx
X= 
= 7.95
= 8.
= Formula = 
2. Calculate the mass percentage of benzene (
) and carbon tetrachloride (
) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans. Mass percentage of 
= 15.28%
Mass percentage of =
=84.72%
Alternatively,
Mass percentage of = (100 – 15.28)%
= 84.72%
3. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans. Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴Mass of carbon tetrachloride = (100 – 30)g
= 70 g
Molar mass of benzene (
) = (6
12 + 6) 
= 78
∴Number of moles of 
= 0.3846 mol
Molar mass of carbon tetrachloride ( ) = 112 + 4355
= 154
∴Number of moles of 
= 0.4545 mol
Thus, the mole fraction of is given as:
= 0.458
4. Calculate the molarity of each of the following solutions: (a) 30 g of
. 
in 4.3 L of solution (b) 30 mL of 0.5 M 
diluted to 500 mL.
Ans. Molarity is given by:
(a) Molar mass of 
= 59 + 2 (14 + 3 16) + 618
= 291
Therefore, Moles of 
= 0.103 mol
Therefore, molarity 
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M 
∴Number of moles present in 30 mL of 0.5 M 
= 0.015 mol
Therefore, molarity
= 0.03 M
5. Calculate (a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is
.
Ans. (a) Molar mass of KI = 39 + 127 = 166
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 =
Therefore, Volume of 100 g solution = 
=
= 83.19 mL
= 
Therefore, molarity of the solution = 
= 1.45 M
(c) Moles of KI = 
Moles of water = 
Therefore, mole fraction of KI = 
= 
= 0.0263
6. 
, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of in water at STP is 0.195 m, calculate Henry’s law constant.
Ans. It is given that the solubility of in water at STP is 0.195 m, i.e., 0.195 mol of is dissolved in 1000 g of water.
Moles of water = 
= 55.56 mol
∴Mole fraction of , x = 
= 
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law:
p= KHx
= 282 bar
7. Henry’s law constant for 
in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of in 500 mL of soda water when packed under 2.5 atm pressure at 298 K.
Ans. It is given that:
KH = 
= 2.5 atm = 
= 
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
Neglectingtheamountofsodapresent
We can write:
500 mL of water = 500 g of water
=
mol of water
= 27.78 mol of water
Now,
Hence, quantity of in 500 mL of soda water = 
= 1.848 g
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.
Ans. 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,
= 
= 180 mm of Hg
= 
= 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
9. Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Ans. (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 nA and nB 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 = 
10. 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
?
Molarity of solution = 
= 16.23 M
Ans. 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 
Then, number of moles of 
= 1.079 mol
Given,
Density of solution = 
Therefore, Volume of 100 g solution = 
= 66.49 mL
= 
Molarity of solution =
= 16.23 M
11. State Henry’s law and mention some important applications?
Ans. 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:
Where,
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 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 of 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
12. What is meant by positive and negative deviations from Raoult’s law and how is the sign of 
related to positive and negative deviations from Raoult’s law?
Ans. According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.
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.
ΔsolH= 0
In the case of solutions showing positive deviations, absorption of heat takes place.
Therefore, = Positive
In the case of solutions showing negative deviations, evolution of heat takes place.
Therefore, = Negative
13. 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:
1. molar mass of the solute
2. vapour pressure of water at 298 K.
Ans. (i) Let, the molar mass of the solute be 
Now, the no. of moles of solvent (water), 
And, the no. of moles of solute, 
p1=2.8kpap1=2.8kpa
Applying the relation:
After the addition of 18 g of water:
Again, applying the relation:
Dividing equation (i) by (ii), we have:
87M + 435 = 84 M +504
3M = 69
M = 23u
Therefore, the molar mass of the solute is
.
(ii) Putting the value of ‘M’ in equation (i), we have:
Hence, the vapour pressure of water at 298 K is 3.53 kPa.
14. 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.
Ans. (i) Phenol 
has the polar group -OH and non-polar group
. Thus, phenol is partially soluble in water.
(ii) Toluene 
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 
has polar -OH group, but it also contains a very bulky non-polar 
group. Thus, pentanol is partially soluble in water.
15. Calculate the depression in the freezing point of water when 10 g of 
is added to 250 g of water. 
.
Ans. Molar mass of 
= 
Therefore, 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.
Therefore, Molality of the solution, 
= 0.3264 
Let a be the degree of dissociation of
undergoes dissociation according to the following equation:
Initial conc. At equilibrium = 
Since α is very small with respect to 1, 1 – α -1
Now, 
= 0.0655
Again,
Initial moles at equilibrium = 
Total moles of equilibrium = 
= 
Therefore, 
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
= 0.65 K
16. Vapor 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.
 | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| acetone /mm Hg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| p chloroform/mm Hg | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
Ans. From the question, we have the following data
| 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| p acetone/mm Hg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| P chloroform/mm Hg | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
| p tota(mm Hg) | 632.8 | 603.0 | 579.5 | 562.1 | 580.4 | 599.5 | 615.3 | 641.8 |
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.
17. 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.
Ans. Molar mass of benzene 
= 
Molar mass of toluene 
= 
Now, no. of moles present in 80 g of benzene = 
= 1.026 mol
And, no. of moles present in 100 g of toluene = 
=1.087 mol
Therefore, Mole fraction of benzene, 
= 0.486
And, mole fraction of toluene, 
= 0.514
It is given that vapour pressure of pure benzene, 
And, vapour pressure of pure toluene, 
Therefore, partial vapour pressure of benzene,
= 24.645 mm Hg
And, partial vapour pressure of toluene, 
= 16.479 mm Hg
Hence, mole fraction of benzene in vapour phase is given by:
= 
= 0.599
= 0.6
18. 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 
and 
respectively, calculate the composition of these gases in water.
Ans. Percentage of oxygen 
in air = 20 %
Percentage of nitrogen 
in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, 
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:
For oxygen:
(Given 
)
For nitrogen:
Hence, the mole fractions of oxygen and nitrogen in water are 
and 
respectively.
19. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of 
in 2 liter of water at 
, assuming that it is completely dissociated.
Ans. When is dissolved in water, 
and 
ions are produced.
Total number of ions produced = 3
Therefore, i =3
Given,
w = 25 mg = 0.025 g
V = 2 L
T = 250C = (25 + 273) K = 298 K
Also, we know that:
R = 0.0821 L atm 
Appling the following relation,
