NCERT solutions for class 12 chemistry chapter 2 Solutions is a resource material of high calibre. chemistry class 12 NCERT solutions chapter 2 pdf can be utilised by the students to prepare for the examination and to solve the NCERT class 12 chemistry chapter 2 exercise solutions along with exemplar problems, MCQ’S, short and long answer questions.
NCERT chemistry class 12 questions solutions chapter 2 given here are very simple and easy to understand. These solutions can help the students to clear all their doubts related to this chapter very easily.
Class 12 Chemistry NCERT Solutions for Solutions Chapter
This chapter holds approximately 5 marks in the board examination. This chapter “solution” teaches about the types of solutions, the concentration of solutions, solubility of solids and gases in a liquid, the vapour pressure of liquid solutions, Raoult’s law, ideal and nonideal solutions, colligative properties and determination of molar masses.
Concepts involved in class 12 chemistry chapter 2 Solutions
 Types of Solutions
 Expressing Concentration of Solutions

Solubility
 The solubility of a Solid in a Liquid
 The solubility of a Gas in a Liquid

Vapour Pressure of Liquid Solutions
 Vapour Pressure of LiquidLiquid Solutions
 Raoult’s Law as a special case of Henry’s Law
 Vapour Pressure of Solutions of Solids in Liquids

Ideal and Nonideal Solutions
 Ideal Solutions
 Nonideal Solutions

Colligative Properties and Determination of Molar Mass
 Relative Lowering of Vapour Pressure
 Elevation of Boiling Point
 Depression of Freezing Point
 Osmosis and Osmotic Pressure
 Abnormal Molar Masses.
NCERT chemistry book includes a chapter on solutions to introduce various important concepts to the students. In this students learn to determine the molarity, molality and mole fraction of solutions, know about Henry’s law constant, mass percentage, etc. The topics provided in the NCERT books are not only important for the class 12 board examination but also for the competitive exams like JEE Mains and NEET. JEE Mains is a national level engineering entrance examination and NEET is a national level examination to take admission in best medical colleges of India.
Solving the NCERT questions will help you to understand the topics in a better and simple way. Sometimes the questions are also asked in the JEE Mains and NEET examinations.
Class 12 Chemistry Solutions Important Questions
Q 2.1) If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl_{4}) and benzene (C_{6}H_{6}).
Answer 2.1:
Mass percentage of Benzene (C_{6}H_{6}) = \(\frac{Mass\; of\; C_{6}H_{6}}{Total \;mass \;of \;the \;solution} v\times 100\)
= \(\frac{Mass\; of\; C_{6}H_{6}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100\)
= \(\frac{22}{22 + 122}\times 100\)
= 15.28%
Mass percentage of Carbon Tetrachloride (CCl_{4}) = \(\frac{Mass\; of\; CCl_{4}}{Total \;mass \;of \;the \;solution} \times 100\)
= \(\frac{Mass\; of\; CCl_{4}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100\)
= \(\frac{122}{22 + 122}\times 100\)
= 84.72%
Q 2.2) If benzene in solution containing 30% by mass in carbon tetrachloride, calculate the mole fraction of benzene?
Answer 2.2:
Assume the mass of benzene be 30 g in the total mass of the solution of 100 g.
Mass of CCl_{4} = (100 − 30) g
= 70 g
Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g \(mol^{1}\)
= 78 g \(mol^{1}\)
Therefore, Number of moles of C_{6}H_{6} = \(\frac{30}{78}\) mol
= 0.3846 mol
Molar mass of CCl_{4} = 1 x 12 + 4 x 355 = 154 g \(mol^{1}\)
Therefore, Number of moles of CCl_{4} = \(\frac{70}{154}\) mol
= 0.4545 mol
Thus, the mole fraction of C_{6}H_{6} is given as:
\(\frac{Number\;of\;moles\;of\;C_{6}H_{6}}{Number\;of\;moles\;of\;C_{6}H_{6} + Number\;of\;moles\;of\;CCl_{4}}\)= \(\frac{0.3846}{0.3846 + 0.4545}\)
= 0.458
Q 2.3) Determine the molarity of each of the solutions given below:
(a) 30 g of Co(NO)_{3}. 6H_{2}O in 4.3 L of solution
(b) 30 mL of 0.5 M H_{2}SO_{4} diluted to 500 mL.
Answer 2.3:
We know that,
Molarity = \(\frac{Moles\;of\;Solute}{Volume\;of\;solution\;in\;litre}\)
(a) Molar mass of Co(NO)_{3}. 6H_{2}O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g \(mol^{1}\)
Therefore, Moles of Co(NO)_{3}. 6H_{2}O = \(\frac{30}{291}\) mol
= 0.103 mol
Therefore, molarity = \(\frac{0.103\; mol}{4.3\; L}\)
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H_{2}SO_{4} = 0.5 mol
Therefore, Number of moles present in 30 mL of 0.5 M H_{2}SO_{4} = \(\frac{0.5\times 30}{1000}\; mol\)
= 0.015 mol
Therefore, molarity = \(\frac{0.015}{0.5\; L}\; mol\)
= 0.03 M
Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH_{2}CONH_{2}) that is required.
Answer 2.4:
Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g \(mol^{1}\)
0.25 molar aqueous solution of urea means:
1000 g of water contains 0.25 mol = (0.25 × 60) g 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 = \(\frac{15\times 2500}{1000 + 15}\; g\)
= 36.95 g
= 37 g of urea (approx.)
Hence, mass of Urea required is 37 g.
Q 2.5) If 1.202 g \(mL^{1}\) is the density of 20% aqueous KI, determine the following:
(a) Molality of KI
(b) Molarity of KI
(c) Mole fraction of KI
Answer 2.5:
(a) Molar mass of KI = 39 + 127 = 166 g \(mol^{1}\)
20% 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 = \(\frac{Moles\;of\;KI}{Mass\;of\;water\;in\;kg}\)
= \(\frac{\frac{20}{166}}{0.08}\; m\)
= 1.506 m
= 1.51 m (approx.)
(b) It is given that the destiny of the solution = 1.202 \(g\; mL^{1}\)
Volume of 100 g solution = \(\frac{Mass}{Density}\)
= \(\frac{100\; g}{1.202\;g \; mL^{1}}\)
= 83.19 mL
= \(83.19\times 10^{3}\; L\)
Therefore, molarity of the solution = \(\frac{\frac{20}{166}\; mol}{83.19\times 10^{3}\; L}\)
= 1.45 M
(c) Moles of KI = \(\frac{20}{166}\) = 0.12 mol
Moles of water = \(\frac{80}{18}\) = 4.44 mol
Therefore, mole = \(\frac{Moles\; of\; KI}{Moles\; of\; KI + Moles\; of \; water}\)
Fraction of KI = \(\frac{0.12}{0.12 + 4.44}\)
= 0.0263
Q 2.6) Calculate Henry’s law constant when the solubility of H_{2}S( a toxic gas with rotten egg like smell) in water at STP is 0.195 m
Answer 2.6:
It is given that the solubility of H_{2}S in water at STP is 0.195 m, i.e., 0.195 mol of H_{2}S is dissolve in 1000 g of water.
Moles of water = \(\frac{1000\; g}{18\; g\; mol^{1}}\)
= 55.56 mol
Therefore, Mole fraction of H_{2}S, x = \(\frac{Moles\;of\;H_{2}S}{Moles\;of\;H_{2}S + Moles\;of\;water}\)
= \(\frac{0.195}{0.195 + 55.56}\)
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law: p = \(K_{H}\)x
=> \(K_{H} = \frac{P}{x}\)
= \(\frac{0.987}{0.0035}\; bar\)
= 282 bar
Q 2.7) Calculate the mass percentage of a solution which is obtained by mixing 300 g of 25 % solution and 400 g of 40 % solution by mass.
Answer 2.7:
Total amount of solute present in the mixture is given by,
\(300\times \frac{25}{100} + 400\times \frac{40}{100}\)= 75 + 160
= 235 g
Total amount of solution = 300 + 400 = 700 g
Therefore, mass percentage of the solute in the resulting solution = \(\frac{235}{700}\times 100\)
= 33.57%
And, mass percentage of the solvent in the resulting solution is:
= (100 – 33.57) %
= 66.43%
Q 2.8) 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.
Answer 2.8:
It is given that:
\(P^{\circ}_{A}\) = 450 mm of Hg \(P^{\circ}_{B}\) = 700 mm of Hg \(P_{total}\) = 600 mm of HgAccording to Raoult’s law:
\(P_{A} = P^{\circ}_{A}x_{A}\) \(P_{B} = P^{\circ}_{B}x_{B} = P^{\circ}_{B}(1 – x_{A})\)Therefore, total pressure, \(P_{total} = P_{A} + P_{B}\)
=> \(P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B}(1 – x_{A})\)
=> \(P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B} – P^{\circ}_{B}x_{A}\)
=> \(P_{total} = (P^{\circ}_{A} – P^{\circ}_{B})x_{A} + P^{\circ}_{B}\)
=> 600 = (450 – 700) x_{A} + 700
=> –100 = –250x_{A}
=> x_{A} = 0.4
Therefore, \(x_{B} = 1 – x_{A}\) = 1 – 0.4 = 0.6
Now, \(P_{A} = P^{\circ}_{A}x_{A}\)
= 450 x 0.4 = 180 mm of Hg
\(P_{B} = P^{\circ}_{B}x_{B}\)= 700 x 0.6 = 420 mm of Hg
Now, in the vapour phase: Mole fraction of liquid A = \(\frac{P_{A}}{P_{A} + P_{B}}\)
= \(\frac{180}{180 + 420}\)
= \(\frac{180}{600}\)
= 0.30
And, mole fraction of liquid B = 1 – 0.30 = 0.70
Q 2.9) Find the vapor pressure of water and and its relative lowering in the solution which is 50 g of urea (NH_{2}CONH_{2}) dissolved in 850 g of water. (Vapor pressure of pure water at 298 K is 23.8 mm Hg)
Answer 2.9:
It is given that vapour pressure of water, \(P^{\circ}_{1}\) = 23.8 mm of Hg
Weight of water taken, \(w_{1}\) = 850 g
Weight of urea taken, \(w_{2}\) = 50 g
Molecular weight of water, \(M_{1}\) = 18 g \(mol^{1}\)
Molecular weight of urea, \(M_{2}\) = 60 g \(mol^{1}\)
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p_{1}.
Now, from Raoult’s law, we have:
\(\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}\)=> \(\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}}\)
=> \(\frac{23.8 – P_{1}}{23.8} = \frac{\frac{50}{60}}{\frac{850}{18} + \frac{50}{60}}\)
=> \(\frac{23.8 – P_{1}}{23.8} = \frac{0.83}{47.22 + 0.83}\)
=> \(\frac{23.8 – P_{1}}{23.8} = 0.0173\)
=> \(P_{1}\) = 23.4 mm of Hg
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.
Q 2.10) How much of sucrose is to be added to 500 g of water such that it boils at 100°C if the molar elevation constant for water is 0.52 K kg mol^{1} and the boiling point of water at 750 mm Hg is 99.63°C.
Answer 2.10:
Here, elevation of boiling point \(\Delta T_{b}\) = (100 + 273) – (99.63 + 273)
= 0.37 K
Mass of water, \(w_{1}\) = 500 g
Molar mass of sucrose (C_{12}H_{22}O_{11}), \(M_{2}\) = 11 x 12 + 22 x 1 + 11 x 16
= 342 g \(mol^{1}\)
Molar elevation constant, K_{b} = 0.52 K kg \(mol^{1}\)
We know that:
\(\Delta T_{b} = \frac{K_{b}\times 1000\times w_{2}}{M_{2}\times w_{1}}\)=> \(w_{2} = \frac{\Delta T_{b}\times M_{2}\times w_{1}}{K_{b}\times 1000}\)
= \(\frac{0.37\times 342\times 500}{0.52\times 1000}\)
= 121.67 g (approximately)
Hence, the amount of sucrose that is to be added is 121.67 g
Q 2.11) To lower the melting point of 75 g of acetic acid by 1.5^{0}C, how much mass of ascorbic acid is needed to be dissolved in the solution where K_{t} = 3.9 K kg \(mol^{1}\).
Answer 2.11:
Mass of acetic acid (w_{1}) = 75 g
Molar mass of ascorbic acid (C_{6}H_{8}O_{6}), M_{2} = 6 x 12 + 8 x 1 + 6 x 16 = 176 g \(mol^{1}\)
Lowering the melting point \(\Delta T_{f}\) = 1.5 K
We know that:
\(\Delta T_{f} = \frac{K_{f}\times w_{2}\times 1000}{M_{2}\times w_{1}}\)=> \(w_{2} = \frac{\Delta T_{f}\times M_{2}\times w_{1}}{K_{f}\times 1000}\)
= \(\frac{1.5\times 176\times 75}{3.9\times 1000}\)
= 5.08 g (approx)
Hence, the amount of ascorbic acid needed to be dissolved is 5.08 g.
Q 2.12) If a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C, calculate the osmotic pressure in Pascal exerted by it?
Answer 2.12:
It is given that:
Volume of water (V) = 450 mL = 0.45 L
Temperature (T) = 37 + 273 = 310 K
Number of moles of the polymer, n = \(\frac{1}{185000}\) mol
We know that:
Osmotic pressure, \(\pi = \frac{n}{V}\; RT\)
= \(\frac{1}{185000}\;mol\times \frac{1}{0.45\;L}\times 8.314\times 10^{3}\; PaL\; K^{1}mol^{1}\times 310\; K\)
= 30.98 Pa
= 31 Pa (approx)
Q 2.13) If the solution contains \(5.00\times 10^{2}\) g of ethane and the partial pressure of ethane over the solution containing \(6.56\times 10^{3}\) g of ethane is 1 bar, then what shall be the partial pressure of the gas?
Answer 2.13:
Molar mass of ethane (C_{2}H_{6}) = 2 x 12 + 6 x 1 = 30 g \(mol^{1}\)
Therefore, number of moles present in \(6.56\times 10^{2}\;g\) of ethane = \(\frac{6.56\times 10^{2}}{30}\)
= \(2.187\times 10^{3}\;mol\)
Let ‘x’ be the number of moles of the solvent, According to Henry’s law,
\(p = K_{H}x\)=> 1 bar = \(K_{H}. \frac{2.187\times 10^{3}}{2.187\times 10^{3} + x}\)
=> 1 bar = \(K_{H}. \frac{2.187\times 10^{3}}{x}\)
=> \(K_{H} = \frac{x}{2.187\times 10^{3}}\) bar (Since x >> \(2.187\times 10^{3}\))
Number of moles present in \(5\times 10^{2}\) g of ethane = \(\frac{5\times 10^{2}}{30}\) mol
= \(1.67\times 10^{3}\;mol\)
According to Henry’s law,
\(p = K_{H}x\)= \(\frac{x}{2.187\times 10^{3}}\times \frac{1.67\times 10^{3}}{(1.67\times 10^{3}) + x}\)
= \(\frac{x}{2.187\times 10^{3}}\times \frac{1.67\times 10^{3}}{x}\) (Since, x>> \(1.67\times 10^{3}\))
= 0.764 bar
Hence, partial pressure of the gas shall be 0.764 bar.
Q 2.14) According to Raoult’s law, what is meant by positive and negative deviaitions and how is the sign of \(\Delta_{sol} H\)related to positive and negative deviations from Raoult’s law?
Answer 2.14:
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 (nonideal 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 twocomponent solution showing positive deviation from Raoult’s law
Vapour pressure of a twocomponent 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.
\(\Delta _{sol}H = 0\)In the case of solutions showing positive deviations, absorption of heat takes place.
\(∴ \Delta _{sol}H = Positive\)In the case of solutions showing negative deviations, evolution of heat takes place.
\(∴ \Delta _{sol}H = Negative\)
Q 2.15) Calculate the molar mass of the nonvolatile solute which is of 2% in an aqueous solution which exerts a pressure of 1.004 bar at the normal boiling point of the solvent.
Answer 2.15:
Vapour pressure of the solution at normal boiling point, \(p_{1}\) = 1.004 bar
Vapour pressure of pure water at normal boiling point, \(p^{\circ}_{1}\) = 1.013 bar
Mass of solute, w_{2} = 2 g
Mass of solvent (water), M_{1} = 18 g \(mol^{1}\)
According to Raoult’s law,
\(\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}\)=> \(\frac{1.013 – 1.004}{1.013} = \frac{2\times 18}{M_{2}\times 98}\)
=> \(\frac{0.009}{1.013} = \frac{2\times 18}{M_{2}\times 98}\)
=> \(M_{2} = \frac{1.013\times 2\times 18}{0.009\times 98}\)
= 41.35 g \(mol^{1}\)
Hence, 41.35 g \(mol^{1}\) is the molar mass of the solute.
Q 2.16) Heptane and octane form an ideal solution. At 373 K, the vapour pressure of the liquid component, heptane is 105.2 kPa and the vapour pressure of the liquid component, octane is 46.8 kPa. Calculate the vaporpressure of the solution when 26.0 g of heptane is mixed with 35 g of octane?
Answer 2.16:
Vapour pressure of heptanes, \(p^{\circ}_{1}\) = 105.2 KPa
Vapour pressure of octane, \(p^{\circ}_{2}\) = 46.8 KPa
We know that,
Molar mass of heptanes (C_{7}H_{16}) = 7 x 12 + 16 x 1 = 100 g \(mol^{1}\)
Therefore, number of moles of heptane = \(\frac{26}{100}\) = 0.26 mol
Molar mass of octane (C_{8}H_{18}) = 8 x 12 + 18 x 1 = 114 g \(mol^{1}\)
Therefore, number of moles of octane = \(\frac{35}{114}\) = 0.31 mol
Mole fraction of heptane, \(x_{1} = \frac{0.26}{0.26 + 0.31}\) = 0.456
And, mole fraction of octane, \(x_{2} = 1 – 0.456\) = 0.544
Now, partial pressure of heptane, \(p_{1} = x_{1}p^{\circ}_{1}\)
= 0.456 x 105.2
= 47.97 KPa
Partial pressure of octane, \(p_{2} = x_{2}p^{\circ}_{2}\)
= 0.544 x 46.8
= 25.46 KPa
Hence, vapour pressure of solution, \(p_{total} = p_{1} + p_{2}\)
= 47.97 + 25.46
= 73.43 KPa
Q 2.17) Calculate vapour pressure of 1 molar solution of a nonvolatile solute in the solution where the vapour pressure of water in it is 12.3 kPa at 300 K.
Answer 2.17:
1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water).
Molar mass of water = 18 g \(mol^{1}\)
Therefore, number of moles present in 1000 g of water = \(\frac{1000}{18}\)
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
\(x_{2} = \frac{1}{1 + 55.56}\) = 0.0177It is given that,
Vapour pressure of water, \(p^{\circ}_{1}\) = 12.3 KPa
Applying the relation, \(\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = x_{2}\)
=> \(\frac{12.3 – p_{1}}{12.3}\) = 0.0177
=> 12.3 – p_{1} = 0.2177
=> p_{1} = 12.0823
= 12.08 KPa (approx)
Hence, the vapour pressure of the solution is 12.08 KPa.
Q 2.18) To reduce the vapor pressure of 114 g of octane to 80%, Calculate the mass of a nonvolatile solute (molar mass 40 g mol−1) which should be dissolved in it
Answer 2.18:
Let \(p^{\circ}_{1}\) be the vapour pressure of pure octane.
Then, after dissolving the nonvolatile solute the vapour pressure of octane is
\(\frac{80}{100}\; p^{\circ}_{1} = 0.8\; p^{\circ}_{1}\)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 x 12 + 18 x 1 = 114 g \(mol^{1}\)
Applying the relation,
\(\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}\)=> \(\frac{p^{\circ}_{1} – 0.8p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times 114}{40\times 114}\)
=> \(\frac{0.2\; p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}}{40}\)
=> 0.2 = \(\frac{w_{2}}{40}\)
=> w_{2} = 8 g
Hence, the required mass of the solute is 8 g.
Q 2.19) Calculate (i) molar mass of the solute in a solution which contains 30 g of nonvolatile solute exactly in 90 g of water which has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Also (ii) Calculate the vapor pressure of water at 298K
Answer 2.19:
(i) Let, the molar mass of the solute be M g \(mol^{1}\)
Now, the number of moles of solvent (water), \(n_{1} = \frac{90\; g}{18\; g\; mol^{1}}\) = 5 mol
And, the number of moles of solute, \(n_{2} = \frac{30\; g}{M\; mol^{1}} = \frac{30}{M}\; mol\)
p_{1} = 2.8 KPa
Applying the relation:
\(\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}\)=> \(\frac{p^{\circ}_{1} – 2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{5 + \frac{30}{M}}\)
=> \(1 – \frac{2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{5M + 30}{M}}\)
=> \(1 – \frac{2.8}{p^{\circ}_{1}} = \frac{30}{5M + 30}\)
=> \(\frac{2.8}{p^{\circ}_{1}} = 1 – \frac{30}{5M + 30}\)
=> \(\frac{2.8}{p^{\circ}_{1}} = \frac{5M + 30 – 30}{5M + 30}\)
=> \(\frac{2.8}{p^{\circ}_{1}} = \frac{5M}{5M + 30}\)
=> \(\frac{p^{\circ}_{1}}{2.8} = \frac{5M + 30}{5M}\) (i)
After the addition of 18 g of water:
\(n_{1} = \frac{90 + 18g}{18} = 6\;mol\) \(p_{1} = 2.9 \;KPa\)Again applying the relation:
\(\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}\)=> \(\frac{p^{\circ}_{1} – 2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{6 + \frac{30}{M}}\)
=> \(1 – \frac{2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{6M + 30}{M}}\)
=> \(1 – \frac{2.9}{p^{\circ}_{1}} = \frac{30}{6M + 30}\)
=> \(\frac{2.9}{p^{\circ}_{1}} = 1 – \frac{30}{6M + 30}\)
=> \(\frac{2.9}{p^{\circ}_{1}} = \frac{6M + 30 – 30}{6M + 30}\)
=> \(\frac{2.9}{p^{\circ}_{1}} = \frac{6M}{6M + 30}\)
=> \(\frac{p^{\circ}_{1}}{2.9} = \frac{6M + 30}{6M}\) (ii)
Dividing equation (i) by (ii), we have:
\(\frac{2.9}{2.8} = \frac{\frac{5M + 30}{5M}}{\frac{6M + 30}{6M}}\)=> \(\frac{2.9}{2.8}\times \frac{6M + 30}{6} = \frac{5M + 30}{5}\)
=> \(2.9\times 5\times (6M + 30) = 2.8\times 6\times (5M + 30)\)
=> 87M + 435 = 84M + 504
=> 3M = 69
=> M = 23 g \(mol^{1}\)
Therefore, 23 g \(mol^{1}\) is the molar mass of the solute.
(ii) Putting the value of ‘M’ in equation (i), we have:
\(\frac{p^{\circ}_{1}}{2.8} = \frac{5\times 23 + 30}{5\times 23}\)=> \(\frac{p^{\circ}_{1}}{2.8} = \frac{145}{115}\)
=> \(p^{\circ}_{1}\) = 3.53KPa
Hence, 3.53 KPa is the vapour pressure of water at 298 K.
Q 2.20) Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K if 5% solution (by mass) of cane sugar in water has freezing point of 271 K.
Answer 2.20:
\(\Delta T_{f}\) = 273.15 – 271 = 2.15 KMolar mass of sugar (C_{12}H_{22}O_{11}) = 12 x 12 + 22 x 1 + 11 x 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 = \(\frac{5}{342}\) mol = 0.0146 mol
Therefore, molality of the solution,
\(m = \frac{0.0146\; mol}{0.095\; kg} = 0.1537 \;mol\;kg^{1}\)Applying the relation,
\(\Delta T_{f} = K_{f}\times m\)=> \(K_{f} = \frac{\Delta T_{f}}{m}\)
= \(\frac{2.15\; K}{0.1537\;mol\;kg^{1}}\)
= 13.99 K kg \(mol^{1}\)
Molar mass of glucose (C_{6}H_{12}O_{6}) = 6 x 12 + 12 x 1 + 6 x 16 = 180 g \(mol^{1}\)
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.
Therefore, number of moles of glucose = \(\frac{5}{180}\) mol = 0.0278 mol
Therefore, molality of the solution, m = \(\frac{0.0278\;mol}{0.095\;kg}\)
= 0.2926 mol \(kg^{1}\)
Applying the relation:
\(\Delta T_{f} = K_{f}\times m\)= \(13.99\;K\;kg\;mol^{1}\times 0.2926\;mol\;kg^{1}\)
= 4.09 K (approx)
Hence, the freezing point of 5 % glucose solution is (273.15 – 4.09) K = 269.06 K.
Q 2.21) Two compounds AB_{2} and AB_{4} are formed from the elements A and B. Calculate atomic masses of A and B, 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}.
Answer 2.21:
We know that,
\(M_{2} = \frac{1000\times w_{2}\times k_{f}}{\Delta T_{f}\times w_{1}}\)Then, \(M_{AB_{2}} = \frac{1000\times 1\times 5.1}{2.3\times 20}\)
= 110.87 g \(mol^{1}\) \(M_{AB_{4}} = \frac{1000\times 1\times 5.1}{1.3\times 20}\)
= 196.15 g \(mol^{1}\)
Now, we have the molar masses of AB_{2} and AB_{4} as 110.87 g \(mol^{1}\) and 196.15 g \(mol^{1}\) respectively.
Let the atomic masses of A and B be x and y respectively.
Now, we can write:
x + 2y = 110.87 …(i)
x + 4y = 196.15 …(ii)
Subtracting equation (i) from (ii), we have
2y = 85.28
=> y = 42.64
Putting the value of ‘y’ in equation (1), we have:
x + 2 (42.64) = 110.87
=> x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.
Q 2.22) At 300 K, the osmotic pressure of 36 g of glucose present in a litre is 4.98 bar. What would be its concentration, if the osmotic pressure of the solution is 1.52 bars at the same temperature?
Answer 2.22:
Given:
T = 300 K
n = 1.52 bar
R = 0.083 bar L \(K^{1}\; mol^{1}\)
Applying the relation, n = CRT
=> C = \(\frac{n}{RT}\)
= \(\frac{1.52\;bar}{0.083\;bar\;L\;K^{1}\;mol^{1}\times 300\;K}\)
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
Q 2.23) In the following pairs, propose the most important type of intermolecular attractive interaction.
(i) nhexane and noctane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).
Answer 2.23:
(i) Van der Wall’s forces of attraction.
(ii) Van der Wall’s forces of attraction.
(iii) Iondiople interaction.
(iv) Dipoledipole interaction.
(v) Dipoledipole interaction.
Q 2.24) Arrange and explain the following in order of increasing solubility in noctane based upon the solutesolvent interactions.
Cyclohexane, KCl, CH_{3}OH, CH_{3}CN.
Answer 2.24:
noctane is a nonpolar solvent. Therefore, the solubility of a nonpolar solute is more than that of a polar solute in the noctane.
The order of increasing polarity is:
Cyclohexane < CH_{3}CN < CH_{3}OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH_{3}OH < CH_{3}CN < Cyclohexane
Q 2.25) Identify which are insoluble, partially soluble and highly soluble in water amongst the following compounds
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer 2.25:
(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 H−bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi)Pentanol (C_{5}H_{11}OH) has polar −OH group, but it also contains a very bulky nonpolar −C5H11 group. Thus, pentanol is partially soluble in water.
Q 2.26) If 92 g of Na^{+} ions are present per kg of water, determine the molality of Na^{+} ions in the lake water which has density of 1.25 g mL^{1}
Answer 2.26:
Number of moles present in 92 g of Na^{+} ions = \(\frac{92\;g}{23\;g\;mol^{1}}\)
= 4 mol
Therefore, molality of Na^{+} ions in the lake = \(\frac{4\;mol}{1\;kg}\)
= 4 m
Q 2.27) If the solubility product of CuS is \(6\times 10^{16}\), determine the maximum molarity of CuS in aqueous solution.
Answer 2.27:
Solubility product of CuS, \(K_{sp} = 6\times 10^{16}\)
Let s be the solubility of CuS in mol L^{1}.
\(CuS \leftrightarrow Cu^{2+} + S^{2}\)Now, s s
\(K_{sp} = [Cu^{2+}] + [S^{2}]\)= s x s
= s^{2}
Then, we have, \(K_{sp} = s^{2} = 6\times 10^{16}\)
=> \(s = \sqrt{6\times 10^{16}}\)
= \(2.45\times 10^{8}\;mol\;L^{1}\)
Hence, \(2.45\times 10^{8}\;mol\;L^{1}\)is the maximum molarity of CuS in an aqueous solution.
Q 2.28) What will the mass percentage of aspirin (C_{9}H_{8}O_{4}) in acetonitrile (CH_{3}CN) be when 6.5 g of C_{9}H_{8}O_{4 }is dissolved in 450 g of CH_{3}CN?
Answer 2.28:
6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.
Then, total mass of the solution = (6.5 + 450) g = 456.5 g
Therefore, mass percentage of C_{9}H_{8}O_{4} = \(\frac{6.5}{456.5}\times 100\)
= 1.424%
Q 2.29) Nalorphene (C_{19}H_{21}NO_{3}), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Calculate the mass of \(1.5\times 10^{3}\; m\) aqueous solution required for the dose of nalorphene generally given is 1.5 mg.
Answer 2.29:
The molar mass of nalorphene (C_{19}H_{21}NO_{3}) = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol^{1}
In \(1.5\times 10^{3}\;m\) aqueous solution of nalorphene,
1 kg (1000 g) of water contains \(1.5\times 10^{3}\;mol\) = \(1.5\times 10^{3}\times 311\) g
= 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, the mass of the solution containing 1.5 mg of nalorphene is:
\(\frac{1000.4665\times 1.5\times 10^{3}}{0.4665}\) g= 3.22 g
Hence, 3.22 g is the required mass of aqueous solution.
Q 2.30) To prepare 250 mL of 0.15 M solution in methanol, determine the amount of benzoic acid (C_{6}H_{5}COOH) that will be required.
Answer 2.30:
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 \(\frac{0.15\times 250}{1000}\) mol of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid (C_{6}H_{5}COOH) = 7 x 12 + 6 x 1 + 2 x 16 = 122 g mol^{1}
Hence, required benzoic acid = 0.0375 mol x 122 g mol^{1} = 4.575 g
Q 2.31) Explain briefly why 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.
Answer 2.31:
Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H^{+} ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid
Q 2.32) When 10 g of CH_{3}CH_{2}CHCICOOH is added to 250 g of water, calculate the depression in the freezing point of water where K_{a} = \(1.4\times 10^{3}\), K_{f} = 1.86 K kg \(mol^{1}\)
Answer 2.32:
Molar mass of CH_{3}CH_{2}CHCICOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1
= 122.5 g mol^{1}
Therefore, No. of moles present in 10 g of CH_{3}CH_{2}CHCICOOH = \(\frac{10\;g}{122.5\;g\;mol^{1}}\)
= 0.0816 mol
It is given that 10 g of CH_{3}CH_{2}CHCICOOH is added to 250 g of water.
Therefore, Molality of the solution, CH_{3}CH_{2}CHCICOOH = \(\frac{0.0186}{250}\times 1000\)
= 0.3264 mol kg^{1}
Let ’a’ be the degree of dissociation of CH_{3}CH_{2}CHCICOOH.
CH_{3}CH_{2}CHCICOOH undergoes dissociation according to the following equation:
\(∴ K_{a} = \frac{C\alpha .C\alpha}{C (1 – \alpha)}\)= \(\frac{C\alpha^{2}}{1 – \alpha}\)
Since a is very small with respect to 1, \(1 – a \approx 1\) \(K_{a} = \frac{C \alpha^{2}}{1}\)
Now,
=> \(K_{a} = C \alpha^{2}\)
=> \(\alpha = \sqrt{\frac{K_{a}}{C}}\)
= \(\sqrt{\frac{1.4\times 10^{3}}{0.3264}}\;\;\;\;(∵ K_{a}=1.4\times 10^{3})\)
= 0.0655
Again,
Total moles of equilibrium = 1 – a + a + a = 1 + a
\(∴ i = \frac{1 + \alpha}{1}\)= \(1 + \alpha\)
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
\(\Delta T_{f} = i.K_{f}m\)= \(1.0655\times 1.86\;K\;kg\;mol^{1}\times 0.3264\;mol\;kg^{1}\)
= 0.65 K
Q 2.33) Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid when 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.
Answer 2.33:
Given:
w_{1} = 500 g
w_{2} = 19.5 g
K_{f} = 1.86 K kg \(mol^{1}\) \(\Delta T_{f}\) = 1 K
We know that:
\(M_{2} = \frac{K_{f}\times w_{2}\times 1000}{\Delta T_{f}\times w_{1}}\)= \(\frac{1.86\;K\;kg\;mol^{1}\times 19.5\;g\times 1000\;g\;kg^{1}}{500\;g\times 1\;K}\)
= \(72.54\;g\;mol^{1}\)
Therefore, observed molar mass of CH_{2}FCOOH, \((M_{2})_{obs} = 72.54\;g\;mol^{1}\)
The calculated molar mass of CH_{2}FCOOH,
\((M_{2})_{cal}\) = 14 + 19 + 12 + 16 + 16 + 1 = 78 g \(mol^{1}\)Therefore, van’t Hoff factor, \(i = \frac{(M_{2})_{cal}}{(M_{2})_{obs}}\) is:
= \(\frac{78\;g\;mol^{1}}{72.54\;g\;mol^{1}}\)
= 1.0753
Let ‘a’ be the degree of dissociation of CH_{2}FCOOH
\(∴ i = \frac{C(1 + \alpha)}{C}\)=> i = 1 + \(\alpha\)
=> \(\alpha\) = i – 1
= 1.0753 – 1
= 0.0753
Now, the value of K_{a} is given as:
\(K_{a} = \frac{[CH_{2}FCOO^{}][H^{+}]}{[CH_{2}FCOOH]}\)= \(\frac{C\alpha.\; C\alpha}{C(1 – \alpha)}\)
= \(\frac{C\alpha^{2}}{1 – \alpha}\)
Taking the volume of the solution as 500 mL, we have the concentration:
C = \(\frac{\frac{19.5}{78}}{500}\times 1000\;M\)
= 0.5 M
Therefore, \(K_{a} = \frac{C\alpha^{2}}{1 – \alpha}\)
= \(\frac{0.5\times (0.0753)^{2}}{1 – 0.0753}\)
= \(\frac{0.5\times 0.00567}{0.9247}\)
= 0.00307 (approx)
= \(3.07\times 10^{3}\)
Q 2.34) Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water when the vapour pressure of water at 293 K is 17.535 mm Hg.
Answer 2.34:
Vapour pressure of water, \(p^{\circ}_{1}\) = 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 x 12 + 12 x 1 + 6 x 16 = 180 g \(mol^{1}\)
Molar mass of water, M_{1} = 18 g \(mol^{1}\)
Then, number of moles of glucose, \(n_{2} = \frac{25}{180\;g\;mol^{1}}\)
= 0.139 mol
And, number of moles of water, \(n_{1} = \frac{450\;g}{18\;g\;mol^{1}}\)
= 25 mol
We know that,
\(\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{1}}{n_{2} + n_{1}}\)=> \(\frac{17.535 – p_{1}}{17.535} = \frac{0.139}{0.139 + 25}\)
=> \(17.535 – p_{1} = \frac{0.139\times 17.535}{25.139}\)
=> \(17.535 – p_{1} = 0.097\)
=> \(p_{1} = 17.44\) mm of Hg
Hence, 17.44 mm of Hg is the vapour pressure of water.
Q 2.35) Calculate the solubility of methane in benzene at 298 K under 760 mm Hg when Henry’s law constant for the molality of methane in benzene at 298 K is \(4.27\times 10^{5}\) mm of Hg.
Answer 2.35:
Given:
p = 760 mm Hg
\(k_{H} = 4.27\times 10^{5}\) mm HgAccording to Henry’s law,
p = k_{H}x
=> x = \(\frac{p}{k_{H}}\)
= \(\frac{760\;mm\;Hg}{4.27\times 10^{5}\;mm\;Hg}\)
= \(177.99\times 10^{5}\)
= \(178\times 10^{5}\) (approx)
Hence, \(178\times 10^{5}\) is the mole fraction of methane in benzene.
Q 2.36) 100 g of liquid A (molar mass 140 g \(mol^{1}\)) was dissolved in 1000 g of liquid B (molar mass180 g \(mol^{1}\)). The vapour pressure of pure liquid B was found to be 500 torr. Find the vapour pressure of pure liquid A and calculate its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer 2.36:
Number of moles of liquid A, \(n_{A} = \frac{100}{140}\) = 0.714 mol
Number of moles of liquid B, \(n_{B} = \frac{1000}{180}\) = 5.556 mol
Then, mole fraction of A, \(x_{A} = \frac{n_{A}}{n_{A} + n_{B}}\)
= \(\frac{0.714}{0.714 + 5.556}\)
= 0.114
And, mole fraction of B, x_{B} = 1 – 0.114 = 0.886
Vapour pressure of pure liquid B, \(p^{\circ}_{B}\) = 500 torr
Therefore, vapour pressure of liquid B in the solution,
\(p_{B} = p^{\circ}_{B}\;x_{B}\)= 500 x 0.886
= 443 torr
Total vapour pressure of the solution, \(p_{total}\) = 475 torr
Therefore, Vapour pressure of liquid A in the solution,
\(p_{A} = p_{total} – p_{B}\)= 475 – 443
= 32 torr
Now, \(p_{A} = p^{\circ}_{A}\;x_{A}\)
=> \(p^{\circ}_{A} = \frac{p_{A}}{x_{A}}\)
= \(\frac{32}{0.114}\)
= 280.7 torr
Hence, 280.7 torr is the vapour pressure of pure liquid A.
Q 2.37) Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot \(p_{total}\), \(p_{chloroform}\) and \(p_{acetone}\) as a function of \(x_{acetone}\). The experimental data observed for different compositions of mixture is.
\(100\times x_{acetone}\)  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 
Answer 2.37:
From the question, we have the following data
\(100\times x_{acetone}\)  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 a negative deviation from the ideal behaviour.
Q 2.38) Benzene and toluene form ideal solution over the entire range of composition. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene where the vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mmHg and 32.06 mm Hg respectively.
Answer 2.38:
Molar mass of benzene (C_{6}H_{6}) = 6 x 12 + 6 x 1 = 78 g \(mol^{1}\)
Molar mass of toluene (C_{6}H_{5}CH_{3}) = 7 x 12 + 8 x 1 = 92 g \(mol^{1}\)
Now, number of moles present in 80 g of benzene = \(\frac{80}{78}\) = 1.026 mol
And, number of moles present in 100 g of toluene = \(\frac{100}{92}\) = 1.087 mol
Therefore, Mole fraction of benzene, \(x_{b} = \frac{1.026}{1.026 + 1.087}\) = 0.486
And, mole fraction of toluene, \(x_{t}\) = 1 – 0.486 = 0.514
It is given that vapour pressure of pure benzene, \(p^{\circ}_{b}\) = 50.71 mm Hg
And, vapour pressure of pure toluene, \(p^{\circ}_{t}\) = 32.06 mm Hg
Therefore, partial pressure of benzene, \(p_{b} = x_{b}\times p^{\circ}_{b}\)
= 0.486 x 50.71
= 24.645 mm Hg
And, partial vapour pressure of toluene, \(p_{t} = x_{t}\times p^{\circ}_{t}\)
= 0.514 x 32.06
= 16.479 mm Hg
Hence, mole fraction of benzene in vapour phase is given by:
\(\frac{p_{b}}{p_{b} + p_{t}}\)= \(\frac{24.645}{24.645 + 16.479}\)
= \(\frac{24.645}{41.124}\)
= 0.599
= 0.6 (approx)
Q 2.39) The air is a mixture of a number of gases. The approximate proportion of the major components, oxygen and nitrogen are 20% is to 79% by volume at 298K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K calculate the composition of oxygen and nitrogen in water if Henry’s law constants for oxygen and nitrogen are \(3.30\times 10^{7}\) mm and \(6.51\times 10^{7}\)mm respectively.
Answer 2.39:
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 (10 x 760) mm Hg = 7600 mm Hg
Therefore,
Partial pressure of oxygen, \(p_{O_{2}} = \frac{20}{100}\times 7600\) mm Hg
= 1520 mm Hg
Partial pressure of nitrogen, \(p_{N_{2}} = \frac{79}{100}\times 7600\)
= 6004 mm Hg
For oxygen:
\(p_{O_{2}} = K_{H}. x_{O_{2}}\)=> \(x_{O_{2}} = \frac{p_{O_{2}}}{K_{H}}\)
= \(\frac{1520\;mm\;Hg}{3.30\times 10^{7}\;mm\;Hg}\;\;\;\;\;\;\;\;\;(Given\;K_{H} = 3.30\times 10^{7}\;mm\;Hg)\)
= \(4.61\times 10^{5}\)
For nitrogen:
\(p_{N_{2}} = K_{H}. x_{N_{2}}\)=> \(x_{N_{2}} = \frac{p_{N_{2}}}{K_{H}}\)
= \(\frac{6004\;mm\;Hg}{6.51\times 10^{7}\;mm\;Hg}\)
= \(9.22\times 10^{5}\)
Hence, \(4.61\times 10^{5}\) and \(9.22\times 10^{5}\) are the mole fractions of oxygen and nitrogen in water.
Q 2.40) To make the osmotic pressure of water as 0.75 atm at 27°C determine the amount of CaCl_{2} (i = 2.47) that is to be dissolved in 2.5 litres of water.
Answer 2.40:
We know that,
\(\pi = i\; \frac{n}{V}\; RT\)=> \(\pi = i\; \frac{w}{MV}\; RT\)
=> \(w = \frac{\pi MV}{iRT}\) \(\pi\) = 0.75 atm
V = 2.5 L
i = 2.47
T = (27 + 273) = 300 K
Here,
R = \(0.0821\;L\;atm\;K^{1}\;mol^{1}\)
M = 1 x 40 + 2 x 35.5
= 111 g \(mol^{1}\)
Therefore, w = \(\frac{0.75\times 111\times 2.5}{2.47\times 0.0821\times 300}\)
= 3.42 g
Hence, 3.42 g is the required amount of CaCl_{2}.
Q 2.41) When 25 mg of K_{2}SO_{4 }is dissolved in 2 litre of water at 25° C assuming that it is completely dissociated, determine the osmotic pressure of the solution.
Answer 2.41:
When K_{2}SO_{4} is dissolved in water, \(K^{+}\; and\; SO_{4}^{2}\) ions are produced.
\(K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2}\)Total number of ions produced = 3
Therefore, i = 3
Given:
w = 25 mg = 0.025 g
V = 2 L
T = 25^{0}C = (25 + 273) = 298 K
Also, we know that:
R = \(0.0821\;L\;atm\;K^{1}\;mol^{1}\)
M = (2 x 39) + (1 x 32) + (4 x 16) = 174 g \(mol^{1}\)
Applying the following relation,
\(\pi = i\; \frac{n}{V}\; RT\)= \(i \frac{w}{M} \frac{1}{v} RT\)
= \(3\times \frac{0.025}{174}\times \frac{1}{2}\times 0.0821\times 298\)
= \(5.27\times 10^{3}\; atm\)
Also Access 
NCERT Exemplar for class 12 Chemistry Chapter 2 
CBSE Notes for class 12 Chemistry Chapter 2 
Students preparing for board examinations are strictly advised to study the chapter thoroughly and should solve the NCERT questions provided at the end of each chapter. The NCERT solutions are designed in such a way that students can understand each and every chapter of the book. Along with NCERT questions students are also advised to solve the CBSE class 12 previous year question papers and sample papers. Solving the previous year questions and sample papers will help you get acquainted with the different types of questions and their marking scheme. Students are always asked to solve the previous year questions like a virtual board examination. Solving the question papers in a stipulated time of 3 hours will help you to know your problemsolving speed and time you take to solve a particular type of question.