Q1 
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. 
Ans: 
A solution is a homogeneous mixtures of two or more than two substances on molecular level.The constitutent of the mixture present in a smaller amount is called the SOLUTE & the one present in larger amount is called the SOLVENT.For eg. Small amount of sugar(solute) dissolved in water(solvent).
SOLUTE + SOLVENT = SOLUTION
There are nine types of solutions formed.They are:
Sno

State of solute

State of solvent

Examples

1

GAS

GAS

Air

2

GAS

LIQUID

Oxygen in water

3

GAS

SOLID

Smole particles in air

4

LIQUID

GAS

Carbon dioxide dissolved in water

5

LIQUID

LIQUID

Alcohol in water

6

LIQUID

SOLID

Mercury in silver

7

SOLID

GAS

Adsorption of hydrogen over palladium

8

SOLID

LIQUID

Sugar in water

9

SOLID

SOLID

Carbon in Iron(steel)

Out of these nine types of solution , solid in liquid, liquid in liquid & gas in liquid are very common.When the components of the solution are mixed,the resulting solution may be in the solid, liquid or gaseous state.They are
(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.
(iii) Solid solutions: The solution in which the solvent is a solid is known as a solid solution.The solute in these solutions may be a gas, liquid or solid. For example, a solution of copper in gold is a solid solution. 

Q2 
Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What kind of solid solution is this likely to be? 
Ans: 
The given solid solution is the example of interstitial solid solution.
For eg tungsten carbide, where tungsten atoms are arranged in a face centered cubic pattern with carbon atoms in octahedral holes. 

Q3 
Define the following terms:
(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage. 
Ans: 
a) Mole Fraction  it is defined as the ratio of number of moles of one component to the total number of moles of solute & solvent present in the solution.
Mole fraction = number of moles of the component / total number of moles of all components
It is denoted by x.
Take it as binary solution, the number of the moles of solute will be n_{A}_{ }and the number of moles of solvent will be n_{B}, then the mole fraction of the solute in the solution is given as:
In the same manner, the mole fraction of the solvent in the solution is given as:
b) Molality it is the number of moles of the solute dissolved per 1000g of the solvent.
It is denoted by m
Molality = mole of solute / mass of solvent in kg
Unit for molality are moles/kg
c) Molarity it is the number of moles of the solute dissolved per litre of the solution.
It is represented by M.
Molarity = moles of solute / volume of solute in litres
Unit for molarity is moles/litre
d) Mass Percentage the mass percentage of a component in a given solution is the mass of the component per 100g of the solution.
Mass percentage of the component = (mass of the component in the solution/total mass of the solution) x100 

Q4 
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}? 
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 (HNO_{3}) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol^{  1}
Then, number of moles of HNO_{3 }= 68 / 63 mol
= 1.08 mol
Also density = 1.504g/mL^{1 } (given)
Therefore from the formula density = mass / volume,we get
Volume of solution = 1000/1.504 = 66.49 mL
Therefore molarity of nitric acid = (1.08/66.49) x 1000 = 16.24 M 

Q5 
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? 
Ans: 
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 = 10 / 180 mol
= 0.056 mol
∴ Molality of solution = 0.056 mol / 0.09kg = 0.62 m
Number of moles of water = 90g / 18g mol^{1} = 5 mol
Mole fraction of glucose (x_{g}) = 0.056 / ( 0.056+5) = 0.011
And, mole fraction of water x_{w }= 1  x_{g}
= 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:
= 100g / 1.2g mL^{1}
= 83.33 mL
=83.33 x 10^{3 }L
∴ Molarity of the solution = 0.056 mol / 83.33 x 10^{3 }L
= 0.67 M 

Q6 
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? 
Ans: 
let the amount of Na_{2}CO_{3} be x
& that of NaHCO_{3} be 1x
Now moles of Na_{2}CO_{3} = x / 106
& moles of NaHCO_{3} = 1x / 84
Now according to question , number of moles of Na_{2}Co_{3} = number of moles of NaHCO_{3}
Therefore x / 106 = 1x / 84
84x = 106106x
84x +106x = 106
190x = 106
Or
x = 106 / 190 = 0.558
Therefore moles of Na_{2}Co_{3} = 0.558 / 106 = 0.00526
&
moles of NaHCO_{3} = 1  0.558 / 84 = 0.0053
Now Hcl reacts with Na_{2}Co_{3} & NaHCO_{3} as follows:
Na_{2}Co_{3} + 2Hcl 2Nacl + H_{2}o + CO_{2}
NaHCO_{3} + Hcl Nacl + H_{2}o + CO_{2}
From the above reactions, 1 mol of Na_{2}Co_{3} will react with 2 mol of Hcl
Therefore 0.00526 mol of Na2Co_{3} will react with 2 x 0.00526 mol of Hcl & similarly 0.00526 mol of NaHCO_{3} will react with 0.00526 mol of Hcl
Total moles of Hcl required to react with mixture of of NaHCO_{3} & Na_{2}Co_{3}
= 2 X 0.00526 + 0.00526 =0.01578 mol
Also according to question 0.1 mol of 0.1 M Hcl is present in 1000 ml
Or
0.01578 mol of 0.1 M Hcl is present in (1000/0.1) x 0.01578 = 158 ml
Hence, 158 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. 

Q7 
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: 
According to question,
300g of 25% solution contains solute = (300 x 25) / 100 = 75g
&
400g of 40% solution contains solute = (400 x 40) / 100 = 160g
Total solute = 75 + 160 = 235g
As given in the question, total solution =300 + 400 = 700g
Percentage of solute in the final solution = (235 x100) / 700 = 33.5%
Percentage of water in the final solution = 100  33.5 = 65.5% 

Q8 
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? 
Ans: 
Calculation of Molality :
Mass of ethylene glycol = 222.6 (Given)
Molar mass of ethylene glycol [C_{2}H_{4}(OH)_{2}]
= 2 X 12 + 6 x 1 + 2 x 16
= 62
Therefore moles of ethylene glycol
= 222.6g / 62 gmol^{1}
= 3.59 mol
Mass of water = 200g (Given)
Therefore molality of the solution is = (moles of ethylene glycol / mass of water) x 1000
= (3.59 / 200) x 1000
= 17.95 m
Calculation of Molarity:
Moles of ethylene glycol = 3.59 mol (already calculated)
Total Mass of solution = 200 + 222.6
= 422.6g
Volume of solution = mass / density volume
= 422.6 / 1.072
= 394.22 ml
now molarity of the solution is = (moles of ethylene glycol / volume of solution) x 1000
= (3.59 / 394.22) x 1000
= 9.11 M 

Q9 
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. 
Ans: 
1) 15 ppm means : 15 parts per million(10^{6}) of the solutions
So, Percent by mass = (mass of chloroform / total mass) x 100
= (15 / 10^{6}) x 100
= 1.5 x 10^{3} %
2) Molality Mass of chloroform = 15 g
Molar mass of chloroform (CHCl_{3}) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5 g mol^{  1}
Moles of chloroform = 15 / 119.5 = 0.1255 mol
Mass of water = 10^{6}
Therefore molality = (moles of chloroform / mass of water ) x 1000
= (0.1255 / 10^{6}) x 1000
= 1.255 x 10^{4} m 

Q10 
What role does the molecular interaction play in a solution of alcohol and water? 
Ans: 
The lower members of alcohols are highly soluble in water but the solubility decreases with increase in the molecular weight. The solubility of lower alcohols in water is due to formation of hydrogen bonds(Hydrogen bonding) between alcohols & water molecules.
However, as the size of alcohol molecule increases, the alkyl groups becomes larger & prevents the formation of hydrogen bonds with water, & hence the solubility goes on decreasing with increase in the length of carbon chain.Also 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. 

Q11 
Why do gases always tend to be less soluble in liquids as the temperature is raised? 
Ans: 
The dissolution of a gas in a liquid is exothermic process.Therefore according to LeChatelier principle,with the increase in temperature,the equilibrium shifts in the backward direction.
Gas + Liquid → Solution + Heat
Therefore the solubility of gas in solution decreases with rise in temperature. 

Q12 
State Henry's law and mention some important applications? 
Ans: 
Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure of the gas in equilibrium with the solution.
Or
It also states that the pressure of a gas over a solution in which the gas is dissolved is proportional to the mole fraction of the gas dissolved in the solution.
The important applications of Henry’s law are as follows:
1) In the production of carbonated beveragesin order to increase the solubility of CO_{2} in cold drinks,beer etc,the bottle are sealed under high pressure.When the bottle is opened under normal atmospheric pressure,the pressure inside the bottle falls to atmospheric pressure & the excess CO_{2} bubbles out of the bottle causing effervescence.
2) At high altitudesthe partial pressure of oxygen at high altitudes is less than the ground level.this results in low concentration of oxygen in the blood & tissues of the peoples
3) In scuba diving during scuba diving,when the diver breaths in compressed air from the supply tank,more nitrogen dissolves in the blood & other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. 

Q13 
The partial pressure of ethane over a solution containing 6.56 x 10^{3} g of ethane is 1 bar. If the solution contains 5.00 x 10^{2} g of ethane, then what shall be the partial pressure of the gas? 
Ans: 
According to Henry’s law
.m = k x p
Substituting the given values in the above equation.
We get
6.56 x10^{3} = k x 1
Or
k = 6.56 x10^{3}
Now when m = 5 x 10^{2},
Then again substituting the given values in Henry’s law equation,we get
5 x 10^{2} = 6.56 x10^{3} x p
Or
p = 7.62 bar 

Q14 
What is meant by positive & negative deviations from raoult’s law & how is the sign of ΔmixH related to positive & negative deviations from Raoult’s law? 
Ans: 
Raoult’s law states that at a given temperature, the vapour pressure of a solution containing non volatile solute is directly proportional the mole fraction of the solvent.
Non ideal solutions shows positive & negative deviations from ideal behavoiur.
Non ideal solutions showing positive deviations from Raoult’s law Consider a binary solution of two components A & B .If the AB interaction in the solutions are weaker than AA & BB interactions in the two liquids forming the solution,then the escaping tendency of A & B types of molecules from the solution becomes more than from pure liquids.
As a result ,each component of solution has a partial vapour pressure greater than expected on the basis of Raoult’s law.This is called positive deviations from Raoult’s law,i.e P_{A}> P_{A} °x_{A } & P_{B} >P_{B}°x_{B}
Δ_{mix}H is positive because energy is required to break AA & BB attractive forces.Hence endothermic process.
Non ideal solutions showing Negative deviations from Raoult’s law in such solutions,the AB interactions are stronger than AA & BB interactions .Due to stronger AB interactions ,the escaping tendency of A & B types of molecules from the solution becomes less than from pure liquids. Consequently, each component of the solution has a partial vapour pressure less than expected on the basis of Raoults law. This is called negative deviations form Raoults law,i.e P_{A}< P_{A} °x_{A } & P_{B} B°x_{B}
Δ_{mix}H is negative because energy is released due to increase in attractive forces. Hence exothermic process. 

Q15 
An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute? 
Ans: 
Here,
Vapour pressure of the solution at normal boiling point (p_{1}) = 1.004 bar (Given)
Vapour pressure of pure water at normal boiling point (p_{1}^{0}) = 1.013 bar
Mass of solute, (w_{2}) = 2 g
Mass of solvent (water), (w_{1}) = 100  2 = 98 g
Molar mass of solvent (water), (M_{1}) = 18 g mol^{  1}
According to Raoult's law,
(p_{1}^{0}  p_{1}) / p_{1}^{0 }= (w_{2 }x M_{1 }) / (M_{2 }x w_{1} )
(1.013  1.004) / 1.013 = (2 x 18) / (M_{2} x 98 )
0.009 / 1.013 = (2 x 18) / (M_{2} x 98 )
M_{2} = (2 x 18 x 1.013) / (0.009 x 98)
M_{2} = 41.35 g mol^{  1}
Hence, the molar mass of the solute is 41.35 g mol^{  1}. 

Q16 
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? 
Ans: 
Vapour pressure of heptane p_{1}^{0} = 105.2 kPa
Vapour pressure of octane p_{2}^{0}= 46.8 kPa
As we know that, Molar mass of heptane (C_{7}H_{16}) = 7 × 12 + 16 × 1 = 100 g mol ^{ 1 }
∴ Number of moles of heptane = 26/100 mol = 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, x_{1} = 0.26 / 0.26 +0.31
= 0.456
And, mole fraction of octane, x_{2} = 1  0.456 = 0.544
Now, partial pressure of heptane, p_{1 }= x_{2 }p_{2}^{0}
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane,p_{2 }= x_{2 }p_{2}^{0}
= 0.544 × 46.8 = 25.46 kPa
Hence, vapour pressure of solution, p_{total }= p1 + p2
= 47.97 + 25.46
= 73.43 kPa 

Q17 
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a nonvolatile solute in it. 
Ans: 
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 = 1000/18
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
x_{2} = 1 / (1+55.56) = 0.0177.
It is given that,
Vapour pressure of water, p_{1}^{0 }= 12.3 kPa
Applying the relation, (P_{1}^{0 } P_{1}) / P_{1}^{0 }= X_{2}
⇒ (12.3  p_{1}) / 12.3 =0.0177
⇒ 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. 

Q18 
Calculate the mass of a nonvolatile solute (molar mass 40 g mol^{1}) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. 
Ans: 
Let the vapour pressure of pure octane be p_{1}^{0}.
Then, the vapour pressure of the octane after dissolving the nonvolatile solute is 80/100 p_{1}^{0} = 0.8 p_{1}^{0}.
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,
(p_{1}^{0}  p_{1}) / p_{1}^{0 }= (w_{2 }x M_{1 }) / (M_{2 }x w_{1} )
⇒ (p_{1}^{0}  0.8 p_{1}^{0}) / p_{1}^{0 }= (w_{2 }x 114_{ }) / (40 _{ }x 114 )
⇒ 0.2 p_{1}^{0} / p_{1}^{0 }= w_{2 }/ 40
⇒ 0.2 = w_{2 }/ 40
⇒ w_{2} = 8 g
Hence, the required mass of the solute is 8 g. 

Q19 
A solution containing 30 g of nonvolatile 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: 
Let, the molar mass of the solute be M g mol ^{ 1}
Now, the no. of moles of solvent (water),n_{1 }= 90g / 18g mol^{1}
And, the no. of moles of solute,n_{2 }= 30g / M mol^{1} = 30 / M mol
p_{1 }= 2.8 kPa
Applying the relation:
(p_{1}^{0}  p_{1}) / p_{1}^{0 }= n_{2 }/ (n_{1 }+ n_{2})
⇒ (p_{1}^{0}  2.8) / p_{1}^{0 }= (30/M)_{ }/ {5_{ }+ (30/M)}
⇒ 1  (2.8/p_{1}^{0}) = (30/M) / {(5M+30)/M}
⇒ 1  (2.8/p_{1}^{0}) = 30 / (5M + 30)
⇒ 2.8/p_{1}^{0 }= 1  30 / (5M + 30)
⇒ 2.8/p_{1}^{0} = (5M + 30  30) / (5M + 30)
⇒ 2.8/p_{1}^{0 }= 5M / (5M+30)
⇒ p_{1}^{0} / 2.8 = (5M+30) / 5M (1)
After the addition of 18 g of water:
n_{1} = (90+18g) / 18 = 6 mol
and the new vapour pressure is p_{1} = 2.9 kPa (Given)
Again, applying the relation:
(p_{1}^{0}  p_{1}) / p_{1}^{0 }= n_{2 }/ (n_{1 }+ n_{2})
⇒ (p_{1}^{0}  2.9) / p_{1}^{0 }= (30/M)_{ }/ {6_{ }+ (30/M)}
⇒ 1  (2.9/p_{1}^{0}) = (30/M) / {(6M+30)/M}
⇒ 1  (2.9/p_{1}^{0}) = 30 / (6M + 30)
⇒ 2.9/p_{1}^{0 }= 1  30 / (6M + 30)
⇒ 2.9/p_{1}^{0} = (6M + 30  30) / (6M + 30)
⇒ 2.9/p_{1}^{0 }= 6M / (6M+30)
⇒ p_{1}^{0} / 2.9 = (6M+30) / 6M (2)
Dividing equation (1) by (2),we get:
2.9 / 2.8 = {(5M+30) / 5M} / {(6M+30) / 6M}
⇒ 2.9 x (6M+30 / 6) = (5M+30 / 5) x 2.8
⇒ 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6
⇒ 87M + 435 = 84M + 504
⇒ 3M = 69
⇒ M = 23u
Therefore, the molar mass of the solute is 23 g mol  1.
(ii) Putting the value of 'M' in equation (i), we get:
⇒ p_{1}^{0} / 2.8 = (5M+30) / 5M
⇒ p_{1}^{0} / 2.8 = (5x23+30) / 5x23
⇒ p_{1}^{0 }= (145 x 2.8) / 115
⇒ p_{1}^{0 }= 3.53
Hence, the vapour pressure of water at 298 K is 3.53 kPa. 

Q20 
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. 
Ans: 
In case of cane sugar:
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 =5/342 mol
= 0.0146 mol
Therefore, molality of the solution,m =0.0146mol / 0.095kg
= 0.1537 kg mol^{  1}
Now applying the relation,
ΔT_{f} = K_{f} × m
⇒ K_{f} = ΔT_{f} / m
⇒ 2.15K / 0.1537 kg mol^{1}
= 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 = 5/180 mol
= 0.0278 mol
Therefore, molality of the solution,m =0.0278 mol / 0.095 kg
= 0.2926 mol kg^{  1}
Applying the relation,
ΔTf = Kf × 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. 

Q21 
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. 
Ans: 
As We know that:
M_{B} = (K_{f} x w_{B} x 1000) / (w_{A} X ΔT_{f})
Now ΔT_{f} = 2.3 , w_{B} = 1.0 , w_{A} = 20, K_{F} = 5.1 (given)
PUTTING THE VALUES IN THE EQUATION
M_{B} = (5.1 x 1 x 1000) / (20 x 2.3) = 110.87 g/mol
Therefore M_{AB2} = 110.9
For AB_{4} compound
ΔTf = 1.3 , w_{b} = 1 ,w_{a} = 20
M_{B} = (5.1 X 1 X 1000) / (20 X 1.3) = 196 g/mol
Therefore M_{AB} = 196
Let x be the atomic mass of A & y be the atomic mass of B,
THEN M_{AB2} = x + 2y = 110.9 (1)
And M_{AB} = x + 4y = 196 (2)
Subtracting 2 from 1 ,we get
2y = 196110.9
y = 85.1 / 2
y = 42.6
Putting the value of y in 1 we get
x = 110.9  2 x 42.6
x = 25.59
Therefore atomic mass of A = 25.59 u Atomic mass of B = 42.6 u. 

Q22 
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 we have given
π_{1}= 4.98
π_{2} = 1.52
C_{1} = 36/180
C_{2} = ? (we have to find)
Now according to van’t hoff equation
Π = CRT
Putting the values in above equation,we get
4.98 = 36/180RT 1
1.52 = c_{2}RT 2
Now dividing equation 2 by 1 ,we get
(c_{2} x 180) / 36 = 1.52 / 4.98
or
c_{2} = 0.0061
Therefore concentration of 2nd solution is 0.0061 M 

Q23 
Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) nhexane and noctane
(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). 
Ans: 
1) both of them has van der waals interactions
2) both of them has van der waals interactions
3) both of them has ion dipole interactions
4) both of them has hydrogen bonding
5) both of them has dipoledipole interactions 

Q24 
Based on solutesolvent interactions, arrange the following in order of increasing solubility in noctane and explain. Cyclohexane, KCl, CH_{3}OH, CH_{3}CN. 
Ans: 
noctane is an organic solvent (liquid)
Out of given examples cyclohexane is strongest organic solvent, so according to like dissolves like, cyclohexane is most likely to be dissolved as a solute in n octane. After cyclohexane, CH_{3}CN
Will be dissolved completely as a solute in n octane, then CH_{3}OH & the least soluble will be Kcl. Also since noctane & cyclohexane, both of them belongs to alkane category, their solubility will be maximum.
Therefore the order of increasing solubility in n octane is as follows:
Kcl < CH_{3}OH < CH_{3}CN < Cyclohexane 

Q25 
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 (C_{6}H_{5}OH) has the polar group OH and nonpolar 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 Hbond with water. Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has polar OH group and can form Hbond. 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 nonpolar C_{5}H_{11} group. Thus, pentanol is partially soluble in water. 

Q26 
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. 
Ans: 
We know molality = moles of solute / mass of solvent in kg
Now mass of Na^{+} ions = 92 g (given)
Moles of Na^{+} ions = 92 / 23 = 4
And mass of water = 1kg
Therefore molality = 4/1 = 4m 

Q27 
If the solubility product of CuS is 6 x 10^{16}, calculate the maximum molarity of CuS in aqueous solution. 
Ans: 
Solubility product of CuS, K_{sp} of CuS = 6 x 10^{16}
If s is the solubility,
then CuS = cu^{2+} + S^{2}
Therefore K_{SP} = { cu^{2+}}{ S^{2}}
Or
K_{SP} = s x s
Or
s = √ K_{SP} = √6 x 10^{16}
= 2.45 x 10^{8} M 

Q28 
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 acetonitrile (CH_{3}CN). 
Ans: 
Mass of aspirin (C_{9}H_{8}O_{4}) = 6.5g (given)
Mass of acetonitrile (CH_{3}CN) = 450g (given)
Now total mass of the solution = 6.5 + 450 = 456.5g
Therefore mass percentage of aspirin (C_{9}H_{8}O_{4}) = (6.5 / 456.5 ) x 100
= 1.42 % 

Q29 
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 x 10^{3}m aqueous solution required for the above dose. 
Ans: 
Molecular mass of nalorphene (C_{19}H_{21}NO_{3}),
= 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g/mol
Moles of nalorphene (C_{19}H_{21}NO_{3}) = 1.5 x 10^{3} x 311 = 4.82 x 10^{6} mol
Now molality = (moles of solute / mass of solvent in g) x 1000
Putting the values in above equation,we get
1.5 x 10^{3} = (4.82 x 10^{6} / mass of water) x 1000
Or
Mass of water = (4.82 x 10^{6} x 1000) / 1.5 x 10^{3}
Therefore mass of water = 3.2g 

Q30 
Calculate the amount of benzoic acid (C_{6}H_{5}COOH) required for preparing 250 mL of 0.15 M solution in methanol. 
Ans: 
We know Molarity = moles of the solute / volume of solution
Putting the given values in above equation,we get
0.15 = (mole of benzoic acid / 250) x 1000
Or
moles of benzoic acid = (0.15 x 250) / 1000
= 0.0375 mol of benzoic acid
Also molecular mass of benzoic acid (C_{6}H_{5}COOH)
= 7 × 12 + 6 × 1 + 2 × 16
= 122 g/mol
Therefore amount of benzoic acid = 0.0375 x 122 = 4.575 g 

Q31 
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. 
Ans: 
The above trend in the depression in the freezing point of water for the same amount of given compounds can be explained on the basis of degree of ionization, which depends upon the strength of the acid.Trifluroacetic acid is more acidic than trichloroacetic acid which is further more acidic than acetic acid.
Therefore the degree of ionization of these acids will decrease in the following order:
Trifluoroactic acid > trichloroacetic acid > acetic acid
Now greater the degree of ionization ,greater will be the depression of freezing point. 

Q32 
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 x 10^{3}, K_{f} = 1.86 K kg mol^{1}. 
Ans: 
Molar mass of CH_{3}CH_{2}CHClCOOH
15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1
= 122.5 g/mol
∴ Moles of CH_{3}CH_{2}CHClCOOH = 10g / 122.5 g/mol
= 0.0816 mol
Therefore molality of the solution
= (0.0816 x 1000) / 250
= 0.3265 mol kg^{1}
Now if a is the degree of dissociation of CH_{3}CH_{2}CHClCOOH,
So, K_{a} = (Cα x Cα) / (C (1α))
K_{a }= Cα^{2 }/ (1α)
Since α is very small with respect to 1, 1  α = 1
K_{a }= Cα^{2 }
α = √K_{α} / C
Putting the values ,We get
α = √1.4 x 10^{3 }/ 0.3265
= 0.0655
Now at equilibrium,the van’t hoff factor i = 1α +α +α/1
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
Therefore ΔT_{f} = i K_{f} m v
= 1.065 v x 1.86 x 0.3265
= 0.647°


Q33 
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. 
Ans: 
Molecular mass of CH_{2}FCOOH
14 + 19 + 12 + 16 + 16 + 1 = 78 g/mol
Now, Moles of CH_{2}FCOOH = 19.5 / 78
= 0.25
Taking the volume of the solution as 500 mL, we have the concentration:
C = (0.25 / 500) X 1000
Therefore Molality = 0.50m
So now putting the value in the formula :
ΔT_{f} = K_{f} x m
=1.86 x 0.50 = 0.93K
Van’t hoff factor = observed freezing point depression / calculated freezing point depression
= 1 / 0.93 = 1.0753
Let α be the degree of dissociation of CH_{2}FCOOH
Now total number of moles = m(1a) + ma +ma = m(1+a)
Or
i = α(1+α) / α = 1 +α = 1.0753
Therefore α = 1.0753 1
= 0.0753
Now the Value of K_{a} is given as:
K_{a }= [CH_{2}FCOO^{}][H^{+}] / CH_{2}FCOOH
= (Cα x Cα) / (C (1α))
= Cα^{2 }/ (1α)
K_{a} = 0.5 X (0.0753)^{2} / (10.0753)
= 0.5 X 0.00567 / 0.09247
= 0.00307 (approx.)
= 3 X 10^{3}


Q34 
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. 
Ans: 
Vapour pressure of water, p_{1}° = 17.535 mm of Hg
Mass of glucose, w_{2} = 25 g
Mass of water, w_{1} = 450 g
We know that,
Molar mass of glucose (C_{6}H_{12}O_{6}),
M_{2 }= 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol ^{ 1}
Molar mass of water, M_{1} = 18 g mol ^{ 1}
Then, number of moles of glucose, n_{1} = 25/180 = 0.139 mol
And, number of moles of water, n_{2} =450/18 = 25 mol
Now, we know that,
(p_{1}°  p°) / p_{1}° = n1 / n_{2} + n_{1}
⇒ 17.535  p° / 17.535 = 0.139 / (0.139+25)
⇒ 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. 

Q35 
Henry's law constant for the molality of methane in benzene at 298 Kis 4.27 x 10^{5} mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg. 
Ans: 
As we know that
p = k x c
We have given p = 760 mm, k = 4.27x 105
Putting the given values in equation
760 = 4.27x 10^{5 }x c
Or
c = 760 / 4.27x 10^{5}
c = 178 x 10^{5} 

Q36 
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. 
Ans: 
Number of Moles of Liquid A, n_{A} = 100 / 140 = 0.714
Number of Moles of Liquid B, n_{B }= 1000 / 180 = 5.556
Then Mole fraction of A = n_{A} / n_{A} + n_{B} = 0.714 / 0.714 + 5.556
= 0.114
Now Mole of fraction of B = 1  0.114 = 0.886
Now p_{total } = p_{A} + p_{B}
Or p_{total} = p°_{A}X_{A} + p°_{B}x_{B}
OR
475 = p°_{A} X 0.114 + 500 X 0.886
OR
p°_{A} = 280.7 torr
Therefore vapour pressure of pure A = 280.7 torr
Vapour pressure of A in solution = 280.7 x 0.114
= 32 torr
Now
p_{A }= p°_{A}X_{A}
Or
p°_{A }= p_{A }/ X_{A}
⇒ 32 / 0.114
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr. 

Q37 
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_{acetoneas} a function of x_{acetone}. The experimental data observed for different compositions of mixture is.
100 ×x_{acetone}

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 

Ans: 
From the question, we have the following data
100 ×x_{acetone}

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_{total}(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. 

Q38 
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. 
Ans: 
Molar mass of benzene(C_{6}H_{6}) = 6 X 12 + 6 X 1 = 78 g/mol
Molar mass of toluene = 7 x 12 + 8 x 1 = 92 g/mol
Now no of moles in 80g of benezen = 80 / 78 = 1.026 mol
No of moles in 100g of toluene = 100 / 92 = 1.087 mol
∴Mole fraction of benzene x_{b} = 1.026 / 1.026 + 1.087 = 0.486
And Mole fraction of toluene,x_{t} = 1  0.486 = 0.514
We have given that
Vapor pressure of pure benzene p_{b}° = 50.71 mm Hg
And, vapour pressure of pure toluene, p_{t}° = 32.06 mm Hg
Therefore partial Vapor pressure of benzene, p_{b} = p_{b} X x_{b}
= 50.71 x 0.486
= 24.65 mm Hg
And partial Vapor pressure of toluene, p_{t} = p_{t} X x_{t}
P_{t} = p°_{t} X x_{t} = 32.06 x 0.514
= 16.48
Total vapour pressure = 24.65 + 16.48 = 41.13 mm Hg
Mole fraction of benzene in vapour phase = 24.65 / 41.13 = 0.60 

Q39 
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 x 10^{7} mm and 6.51 x 10^{7}mm respectively, calculate the composition of these gases in water. 
Ans: 
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, p_{o2 = }20/100 *7600
=1520 mm Hg
Partial pressure of nitrogen,p_{N2 }= 79/100 *7600
= 6004 mmHg
Now, according to Henry's law:
p = K_{H}.x
For oxygen:
p_{o2 }= K_{H}. x_{O2}
⇒x_{O2 }= p_{o2} / K_{H}
= 1520 / 3.30 X 10^{7}
= 4.61 10^{5}
For nitrogen:
p_{N2} = K_{H}.x_{N2}
⇒x_{N2 }= p_{N2} / K_{H}
= 6004 / 6.51 x 10^{7}
= 9.22 x 10^{5}
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10 ^{ 5 }and 9.22 × 10 ^{ 5 }respectively. 

Q40 
Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C. 
Ans: 
We know that
π = i n/V RT
⇒π = i w/MV iRT
⇒ w = πMV / iRT .......................(1)
Now we have given below values:
π = 0.75 atm
V = 2.5L
i = 2.47
T = (27+273) K = 300K
Here,
R = 0.0821L atm k^{1} mol^{1}
M = 1x40 + 2x35.5
= 111 g/mol
Now putting the value in equation 1:
w = 0.75x111x2.5 / 2.47x0.0821x300
=3.42g
Hence, the required amount of CaCl_{2} is 3.42 g.


Q41 
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. 
Ans: 
If K_{2}SO_{4} is completely dissociated then these ions are produced
Then K_{2}SO_{4} = 2K^{+} + SO_{4}^{2}
So total number of ions produced = 3
Therefore I =3
Now molecular mass of K_{2}SO_{4} = 2 x 39 + 1 x 32 + 4 x 16 = 174 g/mol
Now π = I Crt
= I W_{B} X RT / M_{B} X V
= 3 X 25 X 10^{3} X 0.082 X 298 / 174 X 2
= 5.27 X 10^{3} atm. 
