Class 12  Chemistry  Solutions
Question 2.1.
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:
A solution is a homogeneous mixture of two or more than two substances on molecular level.
The constituent 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:
State of Solute 
State of Solvent 
Examples 
1. Gas 
Gas 
Air 
2. Gas 
Liquid 
Oxygen in water 
3. Gas 
Solid 
Smoke particles in air 
4. Liquid 
Gas 
CO_{2} 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. 
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.
Question 2.2.
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?
Answer:
The given solid solution is the example of an interstitial solid solution.
In this solution, particles of solid solute fit into voids or interstices of solvent.
For example: tungsten carbide, where tungsten atoms are arranged in a fcc pattern with carbon atoms in octahedral holes.
Question 2.3.
Define the following terms:
(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage
Answer:
(i) Mole fraction  The mole fraction of a particular component in a solution is the ratio of the
number of moles of that component to the total number of moles of all the components present in the solution.
Mathematically,
Mole Fraction of component
= (Number of moles of given component)/(Total number of moles in a solution)
Mole Fraction is independent of temperature.
(ii) Molality  Molality of a solution is defined as the number of moles of solute dissolved per 1000g [1kg] of the solvent.
It is represented by m.
Molality actually represents the concentration of solution in (mol / kg).
Mathematically,
Molality = (Number of moles of solute)/ (Mass of solvent in Kg)
It is represented by m.
(iii) Molarity The number of moles of solute dissolved per litre of the solution at a particular temperature is called the
molarity of the solution at that temperature.
Molarity actually represents the concentration of a solution in (mol / L).
Mathematically,
Molarity = (Number of moles of solute)/ (Volume of solution in litres)
(iv) Mass percentage  Mass percentage is defined as the mass of the solute in grams dissolved per 100g of the solution.
It is also referred to as weight percentage [w/w].
For example, 10% [by mass] urea solution means that 10 g of urea are present in 100 g of solution, the solvent being only 10010 = 90 g.
Mathematically, the mass percentage of a solute in a solution is given by:
Mass Percentage of Solute =
[(Mass of solute)/ (Mass of solute + Mass of solvent)] x100
Or
Mass Percentage of Solute = [(Mass of solute)/ (Mass of solution)] x (100)
Question 2.4.
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}?
Answer:
Given:
Concentration of Nitric Acid, HNO_{3} = 68%
Density of solution, d = 1.504 g/ml
Molarity = M_{o} = (Number of moles of solute)/ (Volume of solution in litres)
Density, d = (Mass [M])/ (Volume [V])
68% of Nitric acid by mass in aqueous solution means that 68g [[68 × 100]/100] of Nitric acid present in 100g of solution.
⇒ Molecular mass of Nitric Acid, HNO3 = [1 × 1] + [1 × 14] + [16 × 3]
= 63g
⇒Number of moles of Nitric Acid = [68/63]
= 1.079 moles
⇒ Given Density, d = 1.504 g/ml
⇒ Volume, v = [100/1.504]
= 66.489 ml
⇒ Molarity, Mo = [1.079/66.489] × 1000
= 16.23 M
Therefore the molarity of the sample is 16.24 M
Question 2.5.
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?
Answer:
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
Therefore, 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 / 1.2) g mL^{1}
= 83.33 mL
=83.33 x 10^{3} L
Therefore, Molarity of the solution = (0.056 mol / 83.33 x 10^{3}) L
= 0.67 M
Question 2.6.
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?
Answer:
Molar mass of Na_{2}CO_{3} = 106
Molar mass of NaHCO_{3} = 84
1 moles of each makes (106 + 84) g = 190 g
For a equimolar mixture:
190 g of mixture contains 1 moles of each
1 g of mixture contains (1/190) moles of each = 5.26 X 10^{3} mole of each
Now neutralisation reaction is as below:
Na_{2}CO_{3} + 2HCl > 2 NaCl + CO_{2} + H_{2}O
1 mol 2 mol
(1/190) mol (2/190) mol
NaHCO_{3 } + HCl > NaCl + CO_{2} + H_{2}O
1 mol 1 mol
(1/190) mol (1/190) mol
So to neutralise 1 g of mixture HCl required
= ((2/190) + (1/190)) mol = (3/190) mol
1 mol HCl is present in 1 M solution having 1000 ml volume
1 mol HCl is present in 0.1 M solution having (1000/0.1) ml volume
(3/190) mol HCl is present in 0.1 M solution having (1000/0.1) x (3/190) ml volume = 157.89 ml
So 157.89 ml of 0.1M HCl is required.
Question 2.7.
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.
Answer:
Total amount of solute present in the mixture is given by= (300 x (25/100) + 400)
= (75 + 160)
=235 grams
Total amount of solution = (300 + 400) = 700 gms
Therefore, mass percentage (w/w) of the solute in the resulting solution
= [(235/700)] x 100% = 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution
= (100 – 33.57) %
=66.43%
Question 2.8.
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?
Answer:
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
Total Mass of solution = (200 + 222.6)
= 422.6g
Volume of solution = (mass / density volume)
= (422.6 / 1.072)
= 394.22 ml
So the molarity of the solution
= (moles of ethylene glycol / volume of solution) x (1000)
= (3.59 / 394.22) x (1000)
= 9.11 M
Question 2.9.
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.
Answer:
15 ppm means 15 parts of Chloroform in 106 parts of drinking water
Therefore, Mass Percentage = [(Mass of chloroform)/ (total mass)] x (100)
= (15/10^{6}) x (100)
= 1.5 × 10^{3}
Calculation of Molality:
Molecular Mass of Chloroform, CHCl_{3} = [12] + [1] + [35.5 × 3]
=119.5 g
Number of Moles of Chloroform = [15 / 119.5]
= 0.1255 moles
Molality = (Number of moles)/Mas of solvent in kg)
= (0.1255/10^{6}) x (1000)
= 1.255 × 10^{4}
Therefore the Mass Percentage is = 1.5 × 10^{3} and the Molality of the solution
= 1.255 × 10^{4} m.
Question 2.10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
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.
Hydrogen bonding
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.
Question 2.11.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Whenever a gas is dissolved in a liquid, a small amount heat is liberated in the process.
So dissolving a gas in liquid is overall an exothermic process.
So according to the LeChatelier principle, whenever the temperature is increased for a reaction which is exothermic in nature,
the equilibrium shifts backwards and the reaction proceeds in backward direction that means the solution gets dissociated and will give off gas
and hence solubility of gas decreases.
So with the increase in temperature, the solubility of the gases in liquids decreases.
Question 2.12.
State Henry’s law and mention some important applications?
Answer:
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.,
they 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.
Question 2.13.
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?
Answer:
Molar mass of ethane (C_{2}H_{6}) = (2 × 12 + 6 × 1)
= 30 g mol^{−1}
Number of moles present in 6.56 × 10^{−2} 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
1 bar = K_{H} (2.187 x 10^{4})/ (2.187 x 10^{4} +x)
Or 1 bar = K_{H} (2.187 x 10^{4})/(x) (because x>> 2.187 x 10^{4})
Number of moles present in 5.00 × 10^{−2} g of ethane
= (5.00 x 10^{2}) / (30) mol
= 1.67 × 10^{−3} mol
According to Henry’s law,
p = K_{H} x
=(x) / (2.187 x 10^{4}) x (1.67 x 103)/ [(1.67 x 10^{3}) + (x)]
= [(x)/ (2.187 x 10^{4}) x (1.67 x 10^{3})/(x)] (Because x >> 1.67 x 10^{3})
= 7.636 bar
Hence, partial pressure of the gas shall be 7.636 bars.
Question 2.14.
What is meant by positive and negative deviations from Raoult's law and how is the sign of Δ_{mix}H related to positive
and negative deviations from Raoult's law?
Answer:
Raoult’s law states that at a given temperature, the vapour pressure of a solution containing nonvolatile solute
is directly proportional the mole fraction of the solvent.
Nonideal solutions show positive & negative deviations from ideal behaviour.
Nonideal 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}
Question 2.15.
An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bars at the normal boiling point of the solvent.
What is the molar mass of the solute?
Answer:
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}.
Question 2.16.
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?
Answer:
Vapour pressure of heptane p_{1}^{0} = 105.2 kPa
Vapour pressure of octane p_{2}^{0}= 46.8 kPa
Also, Molar mass of heptane (C_{7}H_{16}) = (7 × 12 + 16 × 1) = 100 g mol^{1}
Therefore, 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}
Therefore, 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} = (p_{1} + p_{2})
= (47.97 + 25.46)
= 73.43 kPa
Question 2.17.
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.
Answer:
1 molal solution means 1 mole of solute present in 1000g of water solvent)
Molecular weight of water = H_{2}O = (1 × 2 + 16) = 18g/mol
No. of moles of water, n = (given mass) / (molecular weight)
=> n = (1000/18) = 55.56 gmol^{1}
Mole fraction of solute in solution, x2 = (moles of solute)/ [(moles of solute + moles of water)]
Or x_{2} = (1)/ (1 + 55.56)
⇒ x_{2} = 0.0177
Given vapour pressure of pure water at 300k is 12.3 kPa
From formula:
(p_{1}^{0}  p_{1}) / (p_{1}^{0}) = x_{2}
(12.3  p_{1})/ (12.3) = 0.0177
Or p_{1} = 12.0823 kPa
Itis the vapour pressure of the solution.
Question 2.18.
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%.
Answer:
Let the vapour pressure of pure octane be p10.
Then, the vapour pressure of the octane after dissolving the nonvolatile solute is (80/100) p_{1}^{0} = 0.8 p10.
Molar mass of solute, M_{2} = 40 g mol^{1}
Mass of octane, w1 = 114 g
Molar mass of octane, (C_{8}H_{18}), M1 = (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)
⇒ .2 = (w_{2} / 40)
w_{2} = 8 g
Hence, the required mass of the solute is 8 g.
Question 2.19.
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:
(i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Answer:
Let, the molar mass of the solute be Mg 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 adding 18 g of water:
n_{1} = (90+18g) / (18) = 6 mol
And the new vapour pressure is p_{1} = 2.9 kPa (Given)
Again, using 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.
Question 2.20.
A 5% solution (by mass) of cane sugar in water has freezing point of 271K.
Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.?
Answer:
Given:
Freezing point of solution =271K
Let total mass of solution = 100gram
Then 5% of 100 gram of solution contains 5 g of solute
Now,
Change in freezing point
∆T_{f} = (freezing point of water – freezing point of solution)
= (273.15 – 271)
= 2.15K
Molar mass of glucose (C_{6}H_{12}O_{6}) = 180 g/mol
Molar mass of cane sugar (C_{12}H_{22}O_{11}) =342g/mol
Number of moles = (given mass/molar mass)
Number of moles of (C_{6}H_{12}O_{6}) = (5/180) = 0.028 moles
Number of moles of (C_{12}H_{22}O_{11}) = (5/ 342) = 0.0146 moles
Mass of solvent = (total mass – mass of solute)
= (100 – 5) = 95 g or 0.095Kg
Molality = (number of moles of solute/ mass of solvent in kg)
Molality of cane sugar = (0.0146/0.095) = 0.154m
Molality of C_{6}H_{12}O_{6} (m) = (0.028/0.095) =0.29m
Deviation from freezing point is given by
∆T_{f} = (K_{f} × molality)
Putting values for cane sugar:
2.15= (K_{f} ×0.154)
K_{f} = 13.97
Putting value of glucose
∆T_{f} = (K_{f} × molality)
∆T_{f} = (13.97× 0.29)
∆T_{f} = 4.08K
Now freezing point of 5% glucose in water = (freezing point of water  ∆T_{f}) = (273.15  4.08)
= 269.07K
Question 2.21.
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 K whereas 1.0 g of AB_{4} lowers it by 1.3 K.
The molar depression constant for benzene is 5.1 K kg mol^{–1}.
Calculate atomic masses of A and B.?
Answer:
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
ΔT_{f} = 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,
Therefore, M_{AB2} = (x + 2y) = 110.9  (1)
And M_{AB} = x + 4y = 196  (2)
Subtracting equation 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.
Question 2.22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bars.
If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
Given,
T = 300 K
π = 1.52 bar
R = 0.083 bar L
Applying the relation, π = CRT
Where
π = osmotic pressure of solution
C = concentration of solution
R = universal gas constant
T = temperature
Concentration of the solution is 0.061mol/L
Question 2.23.
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).
Answer:
(i) Both the compounds are nonpolar and they do not attract each other because they do not form any polar ions.
Vanderwaal’s forces of attraction will be dominant in between them as vanderwaal’s forces of attraction are not a result of any chemical or electronic bond.
(ii) Here both the compounds are nonpolar because in I_{2} both the atoms are same so they have same electronegativity and
hence there will be no displacement of electron cloud, it will be in the centre. In case of CCl_{4} molecule,
it has tetrahedral shape so two Cl atoms will cancel the attraction effect from two opposite Cl atoms, hence molecule as a whole is nonpolar.
Therefore they will also have vanderwaal’s forces of attraction.
(iii) NaClO_{4} is ionic in nature as Na, Cl and O all have different electronegativity and their shape is also not symmetric.
So molecular will be ionic in nature and we know that water is polar because O will attract the electron cloud towards it (more electronegative)
hence there will be formation of dipole (two oppositely charged ions separated by a short distance).
Therefore there will be iondipole interaction between them.
(iv) Methanol and acetone are both polar molecules because of the presence of electron withdrawing O atom in methanol and ketone group in acetone.
So they will have dipoledipole interaction.
(v) Acetonitrile is polar compound due to presence of electronegative N atom and acetone due to ketone group.
So, dipoledipole interaction.
Question 2.24.
Based on solutesolvent interactions, arrange the following in order of increasing solubility in noctane and explain.
Cyclohexane, KCl, CH_{3}OH, CH_{3}CN
Answer:
Here noctane is nonpolar solvent due to long chain saturated structure.
We know that “like dissolves like” so a nonpolar compound will be more soluble in nonpolar solvent as compared to polar compound.
So cyclohexane is nonpolar due to symmetric structure. KCl is ionic in nature as it will dissociate into K ^{+} and Cl^{} ions.
CH_{3}CN is polar as mentioned above and CH_{3}OH is also polar in nature.
The order of increasing polarity is:
Cyclohexane < CH_{3}CN < CH_{3}OH < KCl (O is more electronegative than N)
Therefore, the order of increasing solubility is:
KCl < CH_{3}OH < CH_{3}CN < Cyclohexane
Question 2.25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(vi) Pentanol.
Answer:
Water is a polar compound (due to electronegativity difference between O and H). We know that “like dissolves like”.
So, a nonpolar compound will be more soluble in nonpolar solvent as compared to polar compound.
(i) Phenol has the polar group OH and nonpolar group –C_{6}H_{5} and it cannot form H bonding with water (presence of bulky nonpolar 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 Hbond with water.
Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol(OHCH_{2}CH_{2}OH) has polar OH group and can form Hbond with water.
Thus, it is highly soluble in water.
(v) Chloroform is partly soluble as CHCl_{3} is polar in nature due to high electronegativity of Cl atoms, there will be production of partial + charge
on H atom so it can form H bonding with water but it is also surrounded by 3 Cl atoms, so partly soluble.
(vi) Pentanol (C_{5}H_{11}OH) has polar OH group, but it also contains a very bulky nonpolar group –C_{5}H_{11}.
Thus, pentanol is partially soluble in water.
Question 2.26.
If the density of some lake water is 1.25g mL^{–1} and contains 92 g of Na^{+} ions per kg of water, calculate the molality of Na^{+} ions in the lake.
Answer:
Mass of ions = 92g
Molar mass of ions = Na^{+} = 23g (neglect the mass lost due to absence of an electron)
Moles of ions = (mass of ions)/ (molar mass)
n = (92/23) moles
n = 4moles
Molality of solution = (moles of solute)/ (mass of solvent (in kg))
Molality = (4/1) = 4M
Question 2.27.
If the solubility product of CuS is 6 × 10^{–16}, calculate the maximum molarity of CuS in aqueous solution.
Answer:
The Solubility product of CuS (k_{sp}) = 6 × 10^{16}
CuS → Cu ^{2+} + S^{2}
Let the s be solubility of CuS in mol/L
Ksp = [Cu ^{2+}] [S^{2}]
Ksp = solubility product
(6 × 10^{16}) = s × s = s^{2}
s=√ (6 × 10^{16})
S = (2.45 × 10^{8}) mol/L
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10^{8} mol/L
Question 2.28.
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.
Answer:
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 %
Question 2.29.
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.
Answer:
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
Question 2.30.
Calculate the amount of benzoic acid (C_{6}H_{5}COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer:
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
Question 2.31.
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.
Answer:
The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and
trifluoroacetic acid increases in the pattern,
Acetic acid< trichloroacetic acid< trifluoroacetic acid
This is because fluorine is more electronegative than chlorine.
So, trifluoroacetic acid is a stronger acid in comparison to trichloroacetic acid and acetic acid.
And also, acetic acid is the weakest of all.
Explanation: Stronger acid produces more number of ions; therefore it has more ΔT_{f} (depression in freezing point), hence lower freezing point.
As the acidic strength increases, the acid gets more and more ionised.
Trifluoroacetic acid ionizes to the largest extent.
Hence, in this case, trifluoroacetic acid being the strongest acid produces more number of ions (extent of ionisation and concentration of ions are more),
high ΔT_{f} (depression in freezing point) and lower freezing point and vice versa.
Question 2.32.
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 mol^{–1}
Answer:
Molar mass of CH_{3}CH_{2}CHClCOOH
= (15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1)
= 122.5 g/mol
Therefore 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,
CH_{3}CH_{2}CHClCOOH CH_{3}CH_{2}CHClCOO^{} + H^{+}
Initial Concentration C molL^{1 } 0 0
At equilibrium C (1 α) C α C α
So, K_{a} = (C_{α} x C_{α}) / (C (1α))
K_{a} = (C_{α}^{2})/ (1α)
Since α is very small with respect to 1, (1 – α) = 1
K_{a} = C_{α}^{2}
α = √Kα / (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°
Question 2.33.
19.5 g of CH_{2}FCOOH is dissolved in 500 g of water.
The depression in the freezing point of water observed is 1.00 C.
Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Answer:
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:
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
CH_{2}FCOOH CH_{2}FCOOH^{ } + H^{+}
Initial Concentration C molL^{1 } 0 0
At equilibrium C (1 α) C α C α
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}
Question 2.34.
Vapour pressure of water at 293 K is 17.535 mm Hg.
Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:
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
Also,
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
Also, we know that,
(p_{1}°  p°) / (p_{1}°) = (n_{1}) / (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
Question 2.35.
Henry’s law constant for the molality of methane in benzene at 298 K is
4.27 × 10^{5} mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer:
Henry’s law constant K_{H} = 4.27X 10^{5} mm Hg,
p = 760mm Hg,
Using Henry’s law,
p= (K_{H}) x (x_{gas})
x_{gas} = (K_{H}/p)
760 = (4.27x 10^{5}) x (x)
Or
x_{gas} = (760) / (4.27x 10^{5})
x_{gas} = 177.99 x 10^{5}
Question 2.36.
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.
Answer:
Number of moles of liquid A, n_{A} = (100/140) mol
=0.714 mol
Number of moles of liquid B, nB = (100/180) mol
=5.556 mol
Then mole fraction of A, x_{A} = (n_{A})/ (n_{A} + n_{B})
= (0.714)/ (0.714 + 5.556)
=0.114
And mole fraction of B = (x_{B} 1) = (1 0.114)
=0.886
Also,
p_{total} = (p_{A} + p_{B})
= (p^{o}_{A} x_{A}+ p^{o}_{B }x_{B})
475 = (p^{o}_{A }x 0.114) + (500 x0.886)
p^{o}_{A} = 280.7 Torr
Hence, 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^{o}_{A} x_{A}
p^{o}_{A} = (p_{A} /x_{A})
= (32/0.114)
=280.7 Torr
Question 2.37.
Vapour pressures 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 xacetone.
The experimental data observed for different compositions of mixture is:
100 x 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

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
Answer:
100 x 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

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 2.38.
Benzene and toluene form ideal solution over the entire range of composition.
The vapour pressure of pure benzene and naphthalene 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 naphthalene.
Answer:
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 benzene = (80 / 78) = 1.026 mol
No of moles in 100g of toluene = (100 / 92) = 1.087 mol
Therefore, 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
Also 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
Question 2.39.
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 10atm.
At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10^{7} mm and
6.51 × 10^{7} mm respectively, calculate the composition of these gases in water.
Answer:
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_{o}2 = (20/100) x (7600)
=1520 mm Hg
Partial pressure of nitrogen, p_{N}2 = (79/100) x (7600)
= 6004 mmHg
Now, according to Henry's law:
p = K_{H} x
For oxygen:
p_{o}2 = K_{H} x_{O}2
=>x_{O}2 = (p_{o}2 / K_{H})
= (1520 / 3.30) X (10^{7})
= (4.61x 10^{5})
For nitrogen:
p_{N}2 = K_{H} x_{N}2
or x_{N}2 = (p_{N}2 / 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.
Question 2.40.
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.
Answer:
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.
Question 2.41.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K_{2}SO_{4} in 2 litre of water at 25° C,
assuming that it is completely dissociated.
Answer:
Given that
Mass of K_{2}SO_{4}, w = 25 mg = 0.025g (using 1 g = 1000 mg)
Volume V = 2 liter
T = (25 + 273) = 298 K (add 273 to convert in Kelvin)
The reaction of dissociation of K_{2}SO_{4}
K_{2}SO_{4} → 2K^{+} + SO_{4}^{2}
Number if ions produced = (2+ 1) = 3
So van’t Hoff factor i = 3
Use the formula of Osmotic pressure
π = i (n/v) RT
=i (w/M) (1/v) RT
Gas constant, R = 0.0821 L atm K^{1}mol^{1}
Molar mass of K_{2}SO_{4}, M = (2 × 39 + 1 × 32 + 4 × 16) = 174 g mol^{1}
Substituting the values we get
π = (3) x (0.025/174) x (1/2) x (0.0821) x (298)
π= 5.27 × 10^{3} atm
.