NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions

NCERT TEXTBOOK QUESTIONS SOLVED

2.1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: Mass of solution = Mass of C₆H₆ + Mass of CCl₄
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl₄ = 122/144 x 100 = 84.72 %

2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans: 30% by mass of C₆H₆ in CCl₄ => 30 g C₆H₆ in 100 g solution
.’. no. of moles of C₆H₆,(ⁿC₆h₆) = 30/78 = 0.385

2.3.Calculate the molarity of each of the following solutions: (a) 30 g of CO(NO₃)₂.6H₂O in 4.3 L of solution (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.
Ans: (a) Molar mass of CO(NO₃)₂.6H₂O=310.7 g mol^-1
no. of moles = 30/310.7 = 0.0966
Vol. of solution = 4.3 L
Molarity =0.0966/4.3 = 0.022M
(b) 1000 mL of 0.5M  H₂SO₄ contain H₂SO₄ = 0.5 mole
30 mL of 0.5 M H₂SO₄ contain H₂SO₄
=0.5/1000 x 30 = 0.015 mole
Volume of solution = 500mL=0.5 L
Molarity = 0.015/0.5 = 0.03M

2.4.Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH₂CONH₂) = 60 g mol^-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

2.5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.
Ans:

2.6. H₂ S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Ans: Solubility of H₂S gas = 0.195 m
= 0.195 mole in 1 kg of solvent
1 kg of solvent = 1000g

2.7. Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.
Ans.:

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.
Ans:

2.9. Vapour pressure of pure water at 298 K is 23.8 m m Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Ans:

2.10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Ans:

2.11. Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f= 3.9 K kg mol^-1.
Ans:

2.12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Ans:

NCERT EXERCISES

2.1 Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.
Sol. A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.
Types of Solution ExamplesGaseous solutions
(a) Gas in gas Air, mixture of 0₂ and N₂, etc.
(b) Liquid in gas Water vapour
(c) Solid in gas Camphor vapours in N2 gas, smoke etc.Liquid solutions
(a) Gas in liquid C02 dissolved in water (aerated water), and 02 dissolved in water, etc.
(b) Liquid in liquid Ethanol dissolved in water, etc.
(c) Solid in liquid Sugar dissolved in water, saline water, etc.Solid solutions
(a) Gas in solid Solution of hydrogen in palladium
(b) Liquid in solid Amalgams, e.g., Na-Hg
(c) Solid in solid Gold ornaments (Cu/Ag with Au)

2.2 Give an example of a solid solution in which the solute is a gas.
Sol. Solution of hydrogen in palladium and dissolved gases in minerals.

2.3 Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Sol. (i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute

(ii) Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent.
NOTE: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature,(iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.
NOTE: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.(iv) Mass percentage: It is the amount of solute in grams present in 100g of solution.

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?
Sol. 68% nitric acid by mass means that 68g mass of nitric acid is dissolved in 100g mass of solution. Molar mass of HNO₃= 63g mol^-1

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 m L^-1, then what shall be the molarity of the solution?
Sol. 10 percent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol^-1 and molar mass of water = 18g mol^-1

2.6 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂C0₃ and NaHCO₃ containing equimolar amounts of both?
Sol. Calculation of no. of moles of components in the mixture.

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.
Sol.

2.8 An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C₂ H₆O₂ ) and200 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?
Sol.

2.9 A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), 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.
Sol. 15 ppm means 15 parts in million (10⁶) by mass in the solution.

2.10 What role does the molecular interaction play in a solution of alcohol and water?
Sol. Alcohol and water both have strong tendency to form intermolecular hydrogen bonding. On mixing the two, a solution is formed as a result of formation of H-bonds between alcohol and H₂ O molecules but these interactions are weaker and less extensive than those in pure H₂O. Thus they show a positive deviation from ideal behaviour. As a result of this, the solution of alcohol and water will have higher vapour pressure and lower boiling point than that of water and alcohol.

2.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?
Sol. When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.

2.12 State Henry’s law and mention some important applications.
Sol. The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law. It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas Mathematically, P = K_HX where P is the partial pressure of the gas; and X is the mole fraction of the gas in the solution and K_H is Henry’s Law constant.Applications of Henry’s law
(i) In the production of carbonated beverages (as solubility of C0₂ increases at high pressure).
(ii) In the deep sea diving.
(iii) For climbers or people living at high altitudes, where low blood 0₂ causes climbers to become weak and make them unable to think clearly

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?
Sol.

2.14 What is meant by positive and negative deviations from Raoult’s law and how is the sign of AmixH related to positive and negative deviations from Raoult’s law?
Sol. Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆_solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆_sol H is positive. Similarly ∆_solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute.
So there is expansion in volume on solution formation.
Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆_sol H is negative.

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?
Sol.

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 35.0 g of octane?
Sol.

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
Sol. 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.

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%.
Sol.

2.19 A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) molar mass of the solute.
(ii) vapour pressure of water at 298 K.
Sol. Let the molar mass of solute = Mg mol^-1

2.20 A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Sol. Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)

2.21 Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20g 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.
Sol.

2.22 At 300 K, 36g of glucose present in a litre of its solution has an osmotic pressure of 4.08 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Sol.

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) NaCl0₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆0)
Sol. (i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(ii) Both I₂ and CCl₄ are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(iii) NaCl0₄ is an ionic compound and gives Na⁺ and Cl0₄^– ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.
(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
(v) Both CH₃CN and C₃H₆O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

2.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octanc and explain.
Cyclohexane, KCl, CH₃OH, CH₃CN.
Sol. (a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions.
(b) KCl is an ionic compound, KCl will not dissolve in n-octane.
(c) CH₃OH is polar. CH₃OH will dissolve in n-octane.
(d) CH₃CN is polar but lesser than CH₃OH. Therefore, it will dissolve in n-octane but to a greater extent as compared to CH₃OH. Hence, the order is KCl < CH₃OH < CH₃CN < Cyclohexane.

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
Sol. (i) Phenol (having polar – OH group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH) – Partially soluble.

2.26 If the density of some lake water is 1.25 g mL^-1 and contains 92g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.
Sol.

2.27 If the solubility product of CuS is 6 x 10^-16, calculate the maximum molarity of CuS in aqueous solution.
Sol.