NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions provides accurate and adequate chemistry knowledge. Chemistry Class 12 solutions Chapter 2 can be utilised by the students to prepare for the examination and to solve the questions of class 12 chemistry chapter 2 exercise solutions along with exemplar problems, MCQS, short and long answer questions. These solutions help the students to clear all their doubts related to this chapter very easily.

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A solution is a homogeneous mixture of two or more than two components. Solved answers on the chapter solutions are given by our subject experts. These solutions will help the students to understand types of solutions, expressing the concentration of the solution, solubility, ideal and non-ideal solutions, colligative properties and abnormal molar mass. Chemistry class 12 questions and solutions Chapter 2 given here are very simple and easy to understand.

Class 12 NCERT Solutions for Solutions

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 non-ideal solutions, colligative properties and determination of molar masses.

Subtopics for Class 12 Chemistry Chapter 2 – Solutions

  1. Types of Solutions
  2. Expressing Concentration of Solutions
  3. Solubility
    1. The solubility of a Solid in a Liquid
    2. The solubility of a Gas in a Liquid
  4. Vapour Pressure of Liquid Solutions
    1. Vapour Pressure of Liquid-Liquid Solutions
    2. Raoult’s Law as a special case of Henry’s Law
    3. Vapour Pressure of Solutions of Solids in Liquids
  5. Ideal and Nonideal Solutions
    1. Ideal Solutions
    2. Non-ideal Solutions
  6. Colligative Properties and Determination of Molar Mass
    1. Relative Lowering of Vapour Pressure
    2. Elevation of Boiling Point
    3. Depression of Freezing Point
    4. Osmosis and Osmotic Pressure
  7. 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 to the best medical colleges of India.

Solving the NCERT questions will help you to understand the topics in an effective and simple way. Sometimes the questions are also asked in the JEE Mains and NEET examinations.

Class 12 Chemistry NCERT Solutions (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 (CCl4) and benzene (C6H6).

Answer 2.1:

Mass percentage of Benzene (C6H6) = Mass  of  C6H6Total  mass  of  the  solution v×100\frac{Mass\; of\; C_{6}H_{6}}{Total \;mass \;of \;the \;solution}  v\times 100

= Mass  of  C6H6Mass  of  C6H6+Mass  of  CCl4×100\frac{Mass\; of\; C_{6}H_{6}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100

= 2222+122×100\frac{22}{22 + 122}\times 100

= 15.28%

Mass percentage of Carbon Tetrachloride (CCl4) = Mass  of  CCl4Total  mass  of  the  solution×100\frac{Mass\; of\; CCl_{4}}{Total \;mass \;of \;the \;solution} \times 100

= Mass  of  CCl4Mass  of  C6H6+Mass  of  CCl4×100\frac{Mass\; of\; CCl_{4}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100

= 12222+122×100\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 CCl4 = (100 − 30) g

= 70 g

Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol1mol^{-1}

= 78 g mol1mol^{-1}

Therefore, Number of moles of C6H6 = 3078\frac{30}{78} mol

= 0.3846 mol

Molar mass of CCl4 = 1 x 12 + 4 x 355 = 154 g mol1mol^{-1}

 

Therefore, Number of moles of CCl4 = 70154\frac{70}{154} mol

= 0.4545 mol

Thus, the mole fraction of C6H6 is given as:

Number  of  moles  of  C6H6Number  of  moles  of  C6H6+Number  of  moles  of  CCl4\frac{Number\;of\;moles\;of\;C_{6}H_{6}}{Number\;of\;moles\;of\;C_{6}H_{6} + Number\;of\;moles\;of\;CCl_{4}}

= 0.38460.3846+0.4545\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. 6H2O in 4.3 L of solution

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Answer 2.3:

We know that,

Molarity = Moles  of  SoluteVolume  of  solution  in  litre\frac{Moles\;of\;Solute}{Volume\;of\;solution\;in\;litre}

(a) Molar mass of Co(NO)3. 6H2O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g mol1mol^{-1}

Therefore, Moles of Co(NO)3. 6H2O = 30291\frac{30}{291} mol

= 0.103 mol

Therefore, molarity = 0.103  mol4.3  L\frac{0.103\; mol}{4.3\; L}

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

Therefore, Number of moles present in 30 mL of 0.5 M H2SO4 = 0.5×301000  mol\frac{0.5\times 30}{1000}\; mol

= 0.015 mol

Therefore, molarity = 0.0150.5  L  mol\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 (NH2CONH2) that is required.

Answer 2.4:

Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g mol1mol^{-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 = 15×25001000+15  g\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 mL1mL^{-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 mol1mol^{-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 = Moles  of  KIMass  of  water  in  kg\frac{Moles\;of\;KI}{Mass\;of\;water\;in\;kg}

= 201660.08  m\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  mL1g\; mL^{-1}

Volume of 100 g solution = MassDensity\frac{Mass}{Density}

= 100  g1.202  g  mL1\frac{100\; g}{1.202\;g \; mL^{-1}}

= 83.19 mL

= 83.19×103  L83.19\times 10^{-3}\; L

Therefore, molarity of the solution = 20166  mol83.19×103  L\frac{\frac{20}{166}\; mol}{83.19\times 10^{-3}\; L}

= 1.45 M

(c) Moles of KI = 20166\frac{20}{166} = 0.12 mol

Moles of water = 8018\frac{80}{18} = 4.44 mol

Therefore, mole = Moles  of  KIMoles  of  KI+Moles  of  water\frac{Moles\; of\; KI}{Moles\; of\; KI + Moles\; of \; water}

Fraction of KI = 0.120.12+4.44\frac{0.12}{0.12 + 4.44}

= 0.0263

Q 2.6) Calculate Henry’s law constant when the solubility of H2S (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 H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolve in 1000 g of water.

Moles of water = 1000  g18  g  mol1\frac{1000\; g}{18\; g\; mol^{-1}}

= 55.56 mol

Therefore, Mole fraction of H2S, x = Moles  of  H2SMoles  of  H2S+Moles  of  water\frac{Moles\;of\;H_{2}S}{Moles\;of\;H_{2}S + Moles\;of\;water}

= 0.1950.195+55.56\frac{0.195}{0.195 + 55.56}

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henry’s law: p = KHK_{H}x

=> KH=PxK_{H} = \frac{P}{x}

= 0.9870.0035  bar\frac{0.987}{0.0035}\; bar

= 282 bar

Q 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 2.7:

Total amount of solute present in the mixture is given by,

300×25100+400×40100300\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 = 235700×100\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:

PAP^{\circ}_{A} = 450 mm of Hg

PBP^{\circ}_{B} = 700 mm of Hg

PtotalP_{total} = 600 mm of Hg

According to Raoult’s law:

PA=PAxAP_{A} = P^{\circ}_{A}x_{A} PB=PBxB=PB(1xA)P_{B} = P^{\circ}_{B}x_{B} = P^{\circ}_{B}(1 – x_{A})

Therefore, total pressure, Ptotal=PA+PBP_{total} = P_{A} + P_{B}

=> Ptotal=PAxA+PB(1xA)P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B}(1 – x_{A})

=> Ptotal=PAxA+PBPBxAP_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B} – P^{\circ}_{B}x_{A}

=> Ptotal=(PAPB)xA+PBP_{total} = (P^{\circ}_{A} – P^{\circ}_{B})x_{A} + P^{\circ}_{B}

=> 600 = (450 – 700) xA + 700

=> –100 = –250xA

=> xA = 0.4

Therefore, xB=1xAx_{B} = 1 – x_{A} = 1 – 0.4 = 0.6

Now, PA=PAxAP_{A} = P^{\circ}_{A}x_{A}

= 450 x 0.4 = 180 mm of Hg

PB=PBxBP_{B} = P^{\circ}_{B}x_{B}

= 700 x 0.6 = 420 mm of Hg

Now, in the vapour phase: Mole fraction of liquid A = PAPA+PB\frac{P_{A}}{P_{A} + P_{B}}

= 180180+420\frac{180}{180 + 420}

= 180600\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 its relative lowering in the solution which is 50 g of urea (NH2CONH2) 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, P1P^{\circ}_{1} = 23.8 mm of Hg

Weight of water taken, w1w_{1} = 850 g

Weight of urea taken, w2w_{2} = 50 g

Molecular weight of water, M1M_{1} = 18 g mol1mol^{-1}

Molecular weight of urea, M2M_{2} = 60 g mol1mol^{-1}

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.

Now, from Raoult’s law, we have:

P1P1P1=n2n1+n2\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}

=> P1P1P1=w2M2w1M1+w2M2\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}}}

=> 23.8P123.8=506085018+5060\frac{23.8 – P_{1}}{23.8} = \frac{\frac{50}{60}}{\frac{850}{18} + \frac{50}{60}}

=> 23.8P123.8=0.8347.22+0.83\frac{23.8 – P_{1}}{23.8} = \frac{0.83}{47.22 + 0.83}

=> 23.8P123.8=0.0173\frac{23.8 – P_{1}}{23.8} = 0.0173

=> P1P_{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 ΔTb\Delta T_{b} = (100 + 273) – (99.63 + 273)

= 0.37 K

Mass of water, w1w_{1} = 500 g

Molar mass of sucrose (C12H22O11), M2M_{2} = 11 x 12 + 22 x 1 + 11 x 16

= 342 g mol1mol^{-1}

Molar elevation constant, Kb = 0.52 K kg mol1mol^{-1}

We know that:

ΔTb=Kb×1000×w2M2×w1\Delta T_{b} = \frac{K_{b}\times 1000\times w_{2}}{M_{2}\times w_{1}}

=> w2=ΔTb×M2×w1Kb×1000w_{2} = \frac{\Delta T_{b}\times M_{2}\times w_{1}}{K_{b}\times 1000}

= 0.37×342×5000.52×1000\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 1.50C, how much mass of ascorbic acid is needed to be dissolved in the solution where Kt = 3.9 K kg mol1mol^{-1}?

Answer 2.11:

Mass of acetic acid (w1) = 75 g

Molar mass of ascorbic acid (C6H8O6), M2­­ = 6 x 12 + 8 x 1 + 6 x 16 = 176 g mol1mol^{-1}

Lowering the melting point ΔTf\Delta T_{f} = 1.5 K

We know that:

ΔTf=Kf×w2×1000M2×w1\Delta T_{f} = \frac{K_{f}\times w_{2}\times 1000}{M_{2}\times w_{1}}

=> w2=ΔTf×M2×w1Kf×1000w_{2} = \frac{\Delta T_{f}\times M_{2}\times w_{1}}{K_{f}\times 1000}

= 1.5×176×753.9×1000\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 = 1185000\frac{1}{185000} mol

We know that:

Osmotic pressure, π=nV  RT\pi = \frac{n}{V}\; RT

= 1185000  mol×10.45  L×8.314×103  PaL  K1mol1×310  K\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) 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 2.13:

Molar mass of ethane (C2H6) = 2 x 12 + 6 x 1 = 30 g mol1mol^{-1}

Therefore, number of moles present in 6.56×102  g6.56\times 10^{-2}\;g of ethane = 6.56×10230\frac{6.56\times 10^{-2}}{30}

= 2.187×103  mol2.187\times 10^{-3}\;mol

Let ‘x’ be the number of moles of the solvent, According to Henry’s law,

p=KHxp = K_{H}x

=> 1 bar = KH.2.187×1032.187×103+xK_{H}. \frac{2.187\times 10^{-3}}{2.187\times 10^{-3} + x}

=> 1 bar = KH.2.187×103xK_{H}. \frac{2.187\times 10^{-3}}{x}

=> KH=x2.187×103K_{H} = \frac{x}{2.187\times 10^{-3}} bar         (Since x >> 2.187×1032.187\times 10^{-3})

Number of moles present in 5×1025\times 10^{-2} g of ethane = 5×10230\frac{5\times 10^{-2}}{30} mol

= 1.67×103  mol1.67\times 10^{-3}\;mol

According to Henry’s law,

p=KHxp = K_{H}x

= x2.187×103×1.67×103(1.67×103)+x\frac{x}{2.187\times 10^{-3}}\times \frac{1.67\times 10^{-3}}{(1.67\times 10^{-3}) + x}

= x2.187×103×1.67×103x\frac{x}{2.187\times 10^{-3}}\times \frac{1.67\times 10^{-3}}{x}           (Since, x>> 1.67×1031.67\times 10^{-3})

= 0.764 bar

Hence, partial pressure of the gas shall be 0.764 bar.

Q 2.14) What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔsolH\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 (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

1

Vapour pressure of a two-component solution showing positive deviation from Raoult’s law

2

Vapour pressure of a two-component solution showing negative deviation from Raoult’s law

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.

ΔsolH=0\Delta _{sol}H = 0

In the case of solutions showing positive deviations, absorption of heat takes place.

ΔsolH=Positive∴ \Delta _{sol}H = Positive

In the case of solutions showing negative deviations, evolution of heat takes place.

ΔsolH=Negative∴ \Delta _{sol}H = Negative

Q 2.15) An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer 2.15:

Vapour pressure of the solution at normal boiling point, p1p_{1} = 1.004 bar

Vapour pressure of pure water at normal boiling point, p1p^{\circ}_{1} = 1.013 bar

Mass of solute, w2 = 2 g

Mass of solvent (water), M1 = 18 g mol1mol^{-1}

According to Raoult’s law,

p1p1p1=w2×M1M2×w1\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}

=> 1.0131.0041.013=2×18M2×98\frac{1.013 – 1.004}{1.013} = \frac{2\times 18}{M_{2}\times 98}

=> 0.0091.013=2×18M2×98\frac{0.009}{1.013} = \frac{2\times 18}{M_{2}\times 98}

=> M2=1.013×2×180.009×98M_{2} = \frac{1.013\times 2\times 18}{0.009\times 98}

= 41.35 g mol1mol^{-1}

Hence, 41.35 g mol1mol^{-1} is the molar mass of the solute.

Q 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 2.16:

Vapour pressure of heptanes, p1p^{\circ}_{1} = 105.2 KPa

Vapour pressure of octane, p2p^{\circ}_{2} = 46.8 KPa

We know that,

Molar mass of heptanes (C7H16) = 7 x 12 + 16 x 1 = 100 g mol1mol^{-1}

Therefore, number of moles of heptane = 26100\frac{26}{100} = 0.26 mol

Molar mass of octane (C8H18) = 8 x 12 + 18 x 1 = 114 g mol1mol^{-1}

Therefore, number of moles of octane = 35114\frac{35}{114} = 0.31 mol

Mole fraction of heptane, x1=0.260.26+0.31x_{1} = \frac{0.26}{0.26 + 0.31} = 0.456

And, mole fraction of octane, x2=10.456x_{2} = 1 – 0.456 = 0.544

Now, partial pressure of heptane, p1=x1p1p_{1} = x_{1}p^{\circ}_{1}

= 0.456 x 105.2

= 47.97 KPa

Partial pressure of octane, p2=x2p2p_{2} = x_{2}p^{\circ}_{2}

= 0.544 x 46.8

= 25.46 KPa

Hence, vapour pressure of solution, ptotal=p1+p2p_{total} = p_{1} + p_{2}

= 47.97 + 25.46

= 73.43 KPa

Q 2.17) The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

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 mol1mol^{-1}

Therefore, number of moles present in 1000 g of water = 100018\frac{1000}{18}

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

x2=11+55.56x_{2} = \frac{1}{1 + 55.56} = 0.0177

It is given that,

Vapour pressure of water, p1p^{\circ}_{1} = 12.3 KPa

Applying the relation, p1p1p1=x2\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = x_{2}

=> 12.3p112.3\frac{12.3 – p_{1}}{12.3} = 0.0177

=> 12.3 – p1 = 0.2177

=> p1 = 12.0823

= 12.08 KPa (approx)

Hence, the vapour pressure of the solution is 12.08 KPa.

Q 2.18) Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%?

Answer 2.18:

Let p1p^{\circ}_{1} be the vapour pressure of pure octane.

Then, after dissolving the non-volatile solute the vapour pressure of octane is

80100  p1=0.8  p1\frac{80}{100}\; p^{\circ}_{1} = 0.8\; p^{\circ}_{1}

Molar mass of solute, M2 = 40 g mol1mol^{-1}

Mass of octane, w1 = 114 g

Molar mass of octane, (C8H18), M1 = 8 x 12 + 18 x 1 = 114 g mol1mol^{-1}

Applying the relation,

p1p1p1=w2×M1M2×w1\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}

=> p10.8p1p1=w2×11440×114\frac{p^{\circ}_{1} – 0.8p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times 114}{40\times 114}

=> 0.2  p1p1=w240\frac{0.2\; p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}}{40}

=> 0.2 = w240\frac{w_{2}}{40}

=> w2 = 8 g

Hence, the required mass of the solute is 8 g.

Q 2.19) A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

Answer 2.19:

(i) Let, the molar mass of the solute be M g mol1mol^{-1}

Now, the number of moles of solvent (water), n1=90  g18  g  mol1n_{1} = \frac{90\; g}{18\; g\; mol^{-1}} = 5 mol

And, the number of moles of solute, n2=30  gM  mol1=30M  moln_{2} = \frac{30\; g}{M\; mol^{-1}} = \frac{30}{M}\; mol

p1 = 2.8 KPa

Applying the relation:

p1p1p1=n2n1+n2\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}

=> p12.8p1=30M5+30M\frac{p^{\circ}_{1} – 2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{5 + \frac{30}{M}}

=> 12.8p1=30M5M+30M1 – \frac{2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{5M + 30}{M}}

=> 12.8p1=305M+301 – \frac{2.8}{p^{\circ}_{1}} = \frac{30}{5M + 30}

=> 2.8p1=1305M+30\frac{2.8}{p^{\circ}_{1}} = 1 – \frac{30}{5M + 30}

=> 2.8p1=5M+30305M+30\frac{2.8}{p^{\circ}_{1}} = \frac{5M + 30 – 30}{5M + 30}

=> 2.8p1=5M5M+30\frac{2.8}{p^{\circ}_{1}} = \frac{5M}{5M + 30}

=> p12.8=5M+305M\frac{p^{\circ}_{1}}{2.8} = \frac{5M + 30}{5M}           (i)

After the addition of 18 g of water:

n1=90+18g18=6  moln_{1} = \frac{90 + 18g}{18} = 6\;mol p1=2.9  KPap_{1} = 2.9 \;KPa

Again applying the relation:

p1p1p1=n2n1+n2\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}

=> p12.9p1=30M6+30M\frac{p^{\circ}_{1} – 2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{6 + \frac{30}{M}}

=> 12.9p1=30M6M+30M1 – \frac{2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{6M + 30}{M}}

=> 12.9p1=306M+301 – \frac{2.9}{p^{\circ}_{1}} = \frac{30}{6M + 30}

=> 2.9p1=1306M+30\frac{2.9}{p^{\circ}_{1}} = 1 – \frac{30}{6M + 30}

=> 2.9p1=6M+30306M+30\frac{2.9}{p^{\circ}_{1}} = \frac{6M + 30 – 30}{6M + 30}

=> 2.9p1=6M6M+30\frac{2.9}{p^{\circ}_{1}} = \frac{6M}{6M + 30}

=> p12.9=6M+306M\frac{p^{\circ}_{1}}{2.9} = \frac{6M + 30}{6M}           (ii)

Dividing equation (i) by (ii), we have:

2.92.8=5M+305M6M+306M\frac{2.9}{2.8} = \frac{\frac{5M + 30}{5M}}{\frac{6M + 30}{6M}}

=> 2.92.8×6M+306=5M+305\frac{2.9}{2.8}\times \frac{6M + 30}{6} = \frac{5M + 30}{5}

=> 2.9×5×(6M+30)=2.8×6×(5M+30)2.9\times 5\times (6M + 30) = 2.8\times 6\times (5M + 30)

=> 87M + 435 = 84M + 504

=> 3M = 69

=> M = 23 g mol1mol^{-1}

Therefore, 23 g mol1mol^{-1} is the molar mass of the solute.

(ii) Putting the value of ‘M’ in equation (i), we have:

p12.8=5×23+305×23\frac{p^{\circ}_{1}}{2.8} = \frac{5\times 23 + 30}{5\times 23}

=> p12.8=145115\frac{p^{\circ}_{1}}{2.8} = \frac{145}{115}

=> p1p^{\circ}_{1} = 3.53KPa

Hence, 3.53 KPa is the vapour pressure of water at 298 K.

Q 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 2.20:

ΔTf\Delta T_{f} = 273.15 – 271 = 2.15 K

Molar mass of sugar (C12H22O11) = 12 x 12 + 22 x 1 + 11 x 16 = 342 g mol1mol^{-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 = 5342\frac{5}{342} mol = 0.0146 mol

Therefore, molality of the solution,

m=0.0146  mol0.095  kg=0.1537  mol  kg1m = \frac{0.0146\; mol}{0.095\; kg} = 0.1537 \;mol\;kg^{-1}

Applying the relation,

ΔTf=Kf×m\Delta T_{f} = K_{f}\times m

=> Kf=ΔTfmK_{f} = \frac{\Delta T_{f}}{m}

= 2.15  K0.1537  mol  kg1\frac{2.15\; K}{0.1537\;mol\;kg^{-1}}

= 13.99 K kg mol1mol^{-1}

Molar mass of glucose (C6H12O6) = 6 x 12 + 12 x 1 + 6 x 16 = 180 g mol1mol^{-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 = 5180\frac{5}{180} mol = 0.0278 mol

Therefore, molality of the solution, m = 0.0278  mol0.095  kg\frac{0.0278\;mol}{0.095\;kg}

= 0.2926 mol kg1kg^{-1}

Applying the relation:

ΔTf=Kf×m\Delta T_{f} = K_{f}\times m

= 13.99  K  kg  mol1×0.2926  mol  kg113.99\;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 elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 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 2.21:

We know that,

M2=1000×w2×kfΔTf×w1M_{2} = \frac{1000\times w_{2}\times k_{f}}{\Delta T_{f}\times w_{1}}

Then, MAB2=1000×1×5.12.3×20M_{AB_{2}} = \frac{1000\times 1\times 5.1}{2.3\times 20}

= 110.87 g mol1mol^{-1} MAB4=1000×1×5.11.3×20M_{AB_{4}} = \frac{1000\times 1\times 5.1}{1.3\times 20}

= 196.15 g mol1mol^{-1}

Now, we have the molar masses of AB2 and AB4 as 110.87 g mol1mol^{-1} and 196.15 g mol1mol^{-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, 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?

Answer 2.22:

Given:

T = 300 K

n = 1.52 bar

R = 0.083 bar L K1  mol1K^{-1}\; mol^{-1}

Applying the relation, n = CRT

=> C = nRT\frac{n}{RT}

= 1.52  bar0.083  bar  L  K1  mol1×300  K\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) Suggest the most important type of intermolecular attractive interaction in
the following pairs.
(i) n-hexane and n-octane
(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) Ion-dipole interaction.

(iv) Dipole-dipole interaction.

(v) Dipole-dipole interaction.

Q 2.24) Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

Answer 2.24:

n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.

The order of increasing polarity is:

Cyclohexane < CH3CN < CH3OH < KCl

Therefore, the order of increasing solubility is:

KCl < CH3OH < CH3CN < Cyclohexane

Q 2.25) 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.

Answer 2.25:

(i) Phenol (C6H5OH) has the polar group −OH and non-polar group –C6H5. Thus, phenol is partially soluble in water.

(ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water.

Thus, formic acid is highly soluble in water.

(iv)Ethylene glycol has a polar −OH group and can form H−bond. Thus, it is highly soluble in water.

(v) Chloroform is insoluble in water.

(vi)Pentanol (C5H11OH) has a polar −OH group, but it also contains a very bulky nonpolar −C5H11 group. Thus, pentanol is partially soluble in water.

Q 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 molarity of Na+ ions in the lake

Answer 2.26:

Number of moles present in 92 g of Na+ ions = 92  g23  g  mol1\frac{92\;g}{23\;g\;mol^{-1}}

= 4 mol

Therefore, molality of Na+ ions in the lake = 4  mol1  kg\frac{4\;mol}{1\;kg}

= 4 m

Q 2.27) If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution.

Answer 2.27:                                                                   

Solubility product of CuS, Ksp=6×1016K_{sp} = 6\times 10^{-16}

Let s be the solubility of CuS in mol L-1.

CuSCu2++S2CuS \leftrightarrow Cu^{2+} + S^{2-}

Now,                                                   s                  s

Ksp=[Cu2+]+[S2]K_{sp} = [Cu^{2+}] + [S^{2-}]

= s x s

= s2

Then, we have, Ksp=s2=6×1016K_{sp} = s^{2} = 6\times 10^{-16}

=> s=6×1016s = \sqrt{6\times 10^{-16}}

= 2.45×108  mol  L12.45\times 10^{-8}\;mol\;L^{-1}

Hence, 2.45×108  mol  L12.45\times 10^{-8}\;mol\;L^{-1}is the maximum molarity of CuS in an aqueous solution.

Q 2.28) Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Answer 2.28:

6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Then, total mass of the solution = (6.5 + 450) g = 456.5 g

Therefore, mass percentage of C9H8O4 = 6.5456.5×100\frac{6.5}{456.5}\times 100

= 1.424%

Q 2.29) Nalorphene (C19H21NO3), 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 2.29:

The molar mass of nalorphene (C19H21NO3) = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol-1

In 1.5×103  m1.5\times 10^{-3}\;m aqueous solution of nalorphene,

1 kg (1000 g) of water contains 1.5×103  mol1.5\times 10^{-3}\;mol = 1.5×103×3111.5\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:

1000.4665×1.5×1030.4665\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) Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0.15 M solution in methanol.

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 0.15×2501000\frac{0.15\times 250}{1000} mol of benzoic acid

= 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C6H5COOH) = 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) 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 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) Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1

Answer 2.32:

Molar mass of CH3CH2CHCICOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g mol-1

Therefore, No. of moles present in 10 g of CH3CH2CHCICOOH = 10  g122.5  g  mol1\frac{10\;g}{122.5\;g\;mol^{-1}}

= 0.0816 mol

It is given that 10 g of CH3CH2CHCICOOH is added to 250 g of water.

Therefore, Molality of the solution, CH3CH2CHCICOOH = 0.0186250×1000\frac{0.0186}{250}\times 1000

= 0.3264 mol kg-1

Let ’a’ be the degree of dissociation of CH3CH2CHCICOOH.

CH3CH2CHCICOOH undergoes dissociation according to the following equation:

Ka=Cα.CαC(1α)∴ K_{a} = \frac{C\alpha .C\alpha}{C (1 – \alpha)}

= Cα21α\frac{C\alpha^{2}}{1 – \alpha}

Since a is very small with respect to 1, 1a11 – a \approx 1 Ka=Cα21K_{a} = \frac{C \alpha^{2}}{1}

Now,

=> Ka=Cα2K_{a} = C \alpha^{2}

=> α=KaC\alpha = \sqrt{\frac{K_{a}}{C}}

= 1.4×1030.3264        (Ka=1.4×103)\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=1+α1∴ i = \frac{1 + \alpha}{1}

= 1+α1 + \alpha

= 1 + 0.0655

= 1.0655

Hence, the depression in the freezing point of water is given as:

ΔTf=i.Kfm\Delta T_{f} = i.K_{f}m

= 1.0655×1.86  K  kg  mol1×0.3264  mol  kg11.0655\times 1.86\;K\;kg\;mol^{-1}\times 0.3264\;mol\;kg^{-1}

= 0.65 K

Q 2.33) 19.5 g of CH2 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 2.33:

Given:

w1 = 500 g

w = 19.5 g

Kf = 1.86 K kg mol1mol^{-1} ΔTf\Delta T_{f} = 1 K

We know that:

M2=Kf×w2×1000ΔTf×w1M_{2} = \frac{K_{f}\times w_{2}\times 1000}{\Delta T_{f}\times w_{1}}

= 1.86  K  kg  mol1×19.5  g×1000  g  kg1500  g×1  K\frac{1.86\;K\;kg\;mol^{-1}\times 19.5\;g\times 1000\;g\;kg^{-1}}{500\;g\times 1\;K}

= 72.54  g  mol172.54\;g\;mol^{-1}

Therefore, observed molar mass of CH2FCOOH, (M2)obs=72.54  g  mol1(M_{2})_{obs} = 72.54\;g\;mol^{-1}

The calculated molar mass of CH2FCOOH,

(M2)cal(M_{2})_{cal} = 14 + 19 + 12 + 16 + 16 + 1 = 78 g mol1mol^{-1}

Therefore, van’t Hoff factor, i=(M2)cal(M2)obsi = \frac{(M_{2})_{cal}}{(M_{2})_{obs}} is:

= 78  g  mol172.54  g  mol1\frac{78\;g\;mol^{-1}}{72.54\;g\;mol^{-1}}

= 1.0753

Let ‘a’ be the degree of dissociation of CH2FCOOH

i=C(1+α)C∴ i = \frac{C(1 + \alpha)}{C}

=> i = 1 + α\alpha

=> α\alpha = i – 1

= 1.0753 – 1

= 0.0753

Now, the value of Ka is given as:

Ka=[CH2FCOO][H+][CH2FCOOH]K_{a} = \frac{[CH_{2}FCOO^{-}][H^{+}]}{[CH_{2}FCOOH]}

= Cα.  CαC(1α)\frac{C\alpha.\; C\alpha}{C(1 – \alpha)}

= Cα21α\frac{C\alpha^{2}}{1 – \alpha}

Taking the volume of the solution as 500 mL, we have the concentration:

C = 19.578500×1000  M\frac{\frac{19.5}{78}}{500}\times 1000\;M

= 0.5 M

Therefore, Ka=Cα21αK_{a} = \frac{C\alpha^{2}}{1 – \alpha}

= 0.5×(0.0753)210.0753\frac{0.5\times (0.0753)^{2}}{1 – 0.0753}

= 0.5×0.005670.9247\frac{0.5\times 0.00567}{0.9247}

= 0.00307 (approx)

= 3.07×1033.07\times 10^{-3}

Q 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 2.34:

Vapour pressure of water, p1p^{\circ}_{1} = 17.535 mm of Hg

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g

We know that,

Molar mass of glucose (C6H12O6), M2 = 6 x 12 + 12 x 1 + 6 x 16 = 180 g mol1mol^{-1}

Molar mass of water, M1 = 18 g mol1mol^{-1}

Then, number of moles of glucose, n2=25180  g  mol1n_{2} = \frac{25}{180\;g\;mol^{-1}}

= 0.139 mol

And, number of moles of water, n1=450  g18  g  mol1n_{1} = \frac{450\;g}{18\;g\;mol^{-1}}

= 25 mol

We know that,

p1p1p1=n1n2+n1\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{1}}{n_{2} + n_{1}}

=> 17.535p117.535=0.1390.139+25\frac{17.535 – p_{1}}{17.535} = \frac{0.139}{0.139 + 25}

=> 17.535p1=0.139×17.53525.13917.535 – p_{1} = \frac{0.139\times 17.535}{25.139}

=> 17.535p1=0.09717.535 – p_{1} = 0.097