NCERT Solutions for Class 12 Chemistry Chapter 2 - Solutions
NCERT Solutions for Class 12 Chemistry Chapter 2 Free PDF Download
Please Click on Free PDF Download link to Download the NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions
1. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Ans. A solution is a homogeneous mixture of two or more pure substances. Solute and solvent are two components of a solution.
Types of solutions: On the basis of different states of solute and solvent solutions are of three types mainly, which are further categorized into three.
| Type of Solution | Solute | Solvent | Common Examples |
| Gaseous Solutions | Gas | Gas | Mixture of oxygen and nitrogen gases |
| Liquid | Gas | Chloroform mixed with nitrogen gas | |
| Solid | Gas | Camphor in nitrogen gas | |
| Liquid Solutions | Gas | Liquid | Oxygen dissolved in water |
| Liquid | Liquid | Ethanol dissolved in water | |
| Solid | Liquid | Glucose dissolved in water | |
| Solid Solutions | Gas | Solid | Solution of hydrogen in palladium |
| Liquid | Solid | Amalgam of mercury with sodium | |
| Solid | Solid | Copper dissolved in gold |
2. Give an example of a solid solution in which the solute is a gas.
Ans. In case a solid solution is formed between two substances, an interstitial solid solution will be formed. For example: a solution of hydrogen in palladium is a solution in which the solute is a gas.
3. Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Ans. (i) Mole fraction: It is defined as the ratio of the number of moles of the individual component (solute or solvent) to the total number of moles of all components present in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute
$$\text{(X}_\text{A})=\frac{n_{\text{A}}}{n_{\text{A}}+n_{\text{B}}}\\\text{Mole fraction of solvent (X}_\text{B})=\frac{n_{B}}{n_{A}+n_{B}}$$
(ii) Molality: It is defined as the number of moles of a solute present in 1000g (1kg) of a solvent.
$$\text{Molality (m) =}\frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}}=\frac{n}{\text{W}}\\\text{Unit = mol kg}^{\normalsize–1} \text{or m(molal)}\\\textbf{(iii) Molarity:}\space\text{It is defined as the number of moles of solute present in one litre of solution.}\\\text{Molarity (M) =}\frac{\text{Number of moles of solute}}{\text{Volume of solution in litre}}=\frac{n}{\text{v}}$$
Unit = mol L–1 or M(Molar)
(iv) Mass percentage: It is defined as the mass of solute in grams present in 100g of solution.
$$\text{Mass}\space\%\space \text{of solute}=\frac{\text{Mass of solute}}{\text{Mass of solution}}×100$$
4. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504g mL–1 ?
Ans. Given, 68% HNO3 by mass means,
Mass of HNO3 in solution = 68g
Molar mass of HNO3 = 1 + 14 + 16 × 3 = 63g mol–1
Mass of solution = 100g
Density of solution = 1·504g mL–1
$$\text{Volume of solution =}\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{(100\text{g})}{(1.504g \text{mL}^{\normalsize-1})}=66.5\text{mL}=0.0665\text{L}\\\text{Molarity of solution (M) =}\frac{\text{Mass of HNO}_3/\text{Molar mass of HNO}_3}{\text{Volume of solution in Litres}}\\=\frac{(68g/63g\space\text{mol}^{\normalsize-1})}{0.0665\text{L}}=16.23\text{mol}\space \text{L}^{\normalsize-1}=16.23\text{M}$$
Hence, the molarity of solution is 16.23 M.
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?
Ans. Given, 10 per cent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol–1
Molar mass of water = 18g mol–1
$$\text{Moles of glucose =}\frac{10}{180}=0.0555\space\text{moles}\\\text{Moles of water =}\frac{90}{18}=5\text{moles}\\\text{Molality of solution =}\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}=\frac{0.0555}{0.09}=0.62\space\text{m}\\\text{Mole fraction of glucose = X}_\text{g =}\frac{\text{No. of moles of glucose}}{\text{No. of moles of glucose+No. of moles of water}}\\=\frac{0.0555}{5+0.0555}=0.01\\\text{Mole fraction of water = X}_\text{w =}\frac{\text{No. of moles of water}}{\text{xNo. of moles of glucose + No. of moles of water}}\\=\frac{5}{5+0.0555}=0.99\\\text{Density of solution is 1.2g mL}^{\normalsize–1}$$
$$\text{Therefore, Volume of 100g of solution =}\frac{\text{Mass of solution}}{\text{Density}}=\frac{100}{1.2}=83.33\space\text{mL}\\\therefore\space\text{Molarity of solution =}\frac{\text{Number of moles of glucose}}{\text{Volume of solution (L)}}=\frac{0.0555}{83.33}×1000$$
= 0.67 M.
Hence, the molarity of the solution is 0.67M.
6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
Ans. Calculation of number of moles of components in the mixture
Let x g of Na2CO3 is present in the mixture.
∴ (1 – x)g of NaHCO3 is present in the mixture.
Molar mass of Na2CO3 = 2 × 23 + 12 + 3 × 16 = 106g mol–1
and,
Molar mass of NaHCO3 = 23 × 1 + 1 + 12 + 3 × 16 = 84g mol–1
$$\text{No. of moles of Na}_2\text{CO}_3 =\frac{x}{106}\\\text{No. of moles of NaHCO}_3=\frac{(1-x)}{84}$$
As given that the mixture contains equimolar amounts of Na2CO3 and NaHCHO3, therefore
$$\frac{x}{106}=\frac{(1-x)}{84}\\\text{106 – 106x = 84x}\\\text{106 = 190x}\\\therefore\space x=\frac{106}{190}=0.558\text{(g)}\\\therefore\text{No. of moles of Na}_3\text{CO}_3\text{present}=\frac{0.558}{106}=0.00526\\\text{and},\space\text{No. of moles of NaHCO}_3\space\text{present}=\frac{1-0.558}{84}=0.00526\\\text{Calculation of number of moles of HCl required}\\\text{Na}_2\text{CO}_3 + 2\text{HCl}\xrightarrow{}\text{2NaCl + H}_2\text{O}+\text{CO}_{2}\\\text{NaHCO}_3+\text{HCl}\xrightarrow{}\text{NaCl + H}_2\text{O}+\text{CO}_{2}$$
As can be\seen, each mole of Na2CO3 needs 2 moles of HCl,
∴ 0.00526 mole of Na2CO3 needs = 0.00526 × 2 = 0.01052 mole
Each mole of NaHCO3 needs 1 mole of HCl.
∴ 0.00526 mole of NaHCO3 needs = 1 × 0.00526 = 0.00526 mole
Total amount of HCl needed will be = 0.01052 + 0.00526 = 0.01578 mole 0.1 mole of 0.1 M HCl are present in 1000 mL of HCl
$$\therefore\space\text{0.01578 mole of 0.1 M HCl will be present in}=\frac{1000}{0.1}×0.01578=157.8\text{mL}$$
Hence, the volume of 0.1 M HCl required is 157.8 mL.
7. A solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate to mass percentage of the resulting solution.
$$\textbf{Ans.}\space\text{Mass of one component in solution =}\frac{(300 g)×25}{100}=75\text{g}\\\text{Mass of other component in solution}=\frac{(400 \text{g})×25}{100}=160 \text{g}\\\text{Total mass of both components = (75 + 160)g = 235g}\\\text{Total mass of solution = (300 + 400)g = 700g}\\\%\text{of solute in the final solution =}\frac{(235g)}{(700g)}×100=33.57\%$$
% of solvent in the final solution = 100 – 33.57 = 66.43%
Hence, mass percentage of the resulting solution ia 66.43%.
8. An antifreeze solution is prepared from 222.6g of ethylene glycol, (C2H6O2) and 200g of water. Calculate the molality of the solution. If the density of the solution is 1.072g mL–1, then what shall be the molarity of the solution?
Ans. Given, Mass of solute = 222.6g
Molar mass of solute, C2H4(OH)2 = 12 × 2 + 4 + 2 (12 + 1) = 62g mol–1
$$\therefore\space\text{Moles of solute =}\frac{222.6}{62}=3.59\\\text{Mass of solvent = 200g}\\\therefore\space\text{Molality =}\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}=\frac{3.59}{200}×1000=17.95\space\text{mol kg}^{\normalsize-1}\\\text{Density of solution = 1.072g/mL}\\\text{ Total mass of solution = 422.6g}\\\text{Volume of solution =}\frac{\text{Mass}}{\text{Density}}=\frac{422.6}{1.072}=394.21\text{mL.}\\\therefore\space\text{Molarity =}\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}=\frac{3.59}{394.2}×1000=9.1\text{mol L}^{\normalsize-1}$$
Hence, the molarity of the solution is 9.1 mol L–1.
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.
Ans. (i) 15 ppm means 15 parts in million (106) by mass in the solution.
$$\therefore\space\text{Percentage by mass of CHCl}_3=\frac{15}{10^{6}}×100=15×10^{\normalsize -14}\%=1.5×10^{\normalsize-3}\%$$
(ii) As only 15g of chloroform is present in 106g of the solution, mass of the solvent = 106g = 103 kg
Molar mass of CHCl3 = 12 + 1 + 3 × 35.5 = 199.5g mol–1
$$\text{Moles of CHCl}_3=\frac{\text{Mass}}{\text{Molar Mass}}=\frac{15}{119.5}=0.125\space\text{moles}\\\therefore\space\text{Molality =}\frac{\text{No. of moles of CHCl}_3}{\text{Mass of solvent (kg)}}=\frac{0.125}{10^{3}}=1.25×10^{\normalsize-3}\text{m}$$
10. What role does the molecular interaction play in solution of alcohol in water?
Ans. In pure water, the molecules are held tightly by strong H-bond. When alcohol and water are mixed together, the alcohol and water also interact with each other by H-bond. But the H-bond formed between alcohol and water is weaker than that of pure water and alcohol. This will lead to increase in vapour pressure of the solution and also decrease in boiling point.
11. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Ans. When gases are dissolved in liquid, 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 as the temperature is raised.
12. State Henry’s law and mention some of its important applications.
Ans. Henry’s law states that:
“The solubility of a gas in a liquid is directly proportional to the pressure of the gas at a constant temperature.”
Or
“The partial pressure of a gas in v
apour phase (p) is proportional to the mole fraction of the gas (x) in the solution”.
It is expressed as, p = KHx
Where, KH is Henry’s law constant.
Applications of Henry’s law:
(i) In packing of soda cans: In order to increase the solubility of CO2 gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e., water. As a result, it will rush out of the bottle with a fizz or produce a hissing noise.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimise the painful effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.
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?
Ans. Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1 = 30 g mol–1
$$\therefore\space\text{Number of moles present in 6.56 × 10}^{\normalsize–3} \text{g of ethane =}\frac{6.56×10^{\normalsize-3}}{30}$$
= 2.187 × 10–4 mol
Let the number of moles of the solv
ent be x.
According to Henry’s law,
p = KHx
$$\Rarr\space\text{1 bar}=\text{K}_\text{H}\frac{2.187×10^{\normalsize-4}}{2.187×10^{\normalsize-4}+x}\\\Rarr\space1\text{bar}=\text{K}_\text{H}\frac{2.187×10^{\normalsize-4}}{x}(\text{since }x>> 2.187×10^{\normalsize-4})\\\text{K}_\text{H}=\frac{x}{2.187}×10^{\normalsize-4}\text{bar}\\\text{Number of moles present in 5.00 × 10}^{\normalsize–2}\text{g of ethane =}\frac{5.00×10^{\normalsize-2}}{30}\text{mol}$$
= 1.67 × 10–3 mol
According to Henry’ s law,
p = KHx
$$=\frac{x}{2.187×10^{4}}×\frac{1.67×10^{\normalsize-3}}{1.67×10^{\normalsize-3}+x}\\\Rarr\space\frac{x}{2.187×10^{\normalsize-4}}=\frac{1.67×10^{\normalsize-3}}{x}(\text{since x>>1.67×10}^{\normalsize-3})$$
= 7.636 bar
Hence, partial pressure of the gas shall be 7.636 bar .
14. What is meant by positive and negative deviaitions from Raoult’s law and how is the sign of ΔmixH related to positive and negative deviations from Raoult’s law?
Ans. Positive deviation: Positive deviation from Raoult’s law occurs when the total vapour pressure of the solution is more than corresponding vapour pressure in case of ideal solution. In such case, intermolecular interaction between solute and solvent are weaker than solute-solute and solvent-solvent interaction.It leads to increase in vapour pressure resulting in positive deviation.
Negative deviation: Negative deviation from Raoult’s law occurs when the total vapour pressure of the solution is less than corresponding vapour pressure in case of the ideal solution. In such case, intermolecular interaction between solute-solute and solvent-solvent are weaker than that of solute-solvent interaction. It leads to decrease in vapour pressure resulting in negative deviation.
In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.
ΔmixH = 0
In the case of solutions showing positive deviations, absorption of heat takes place.
∴ ΔmixH = Positive
In the case of solutions showing negative deviations, evolution of heat takes place.
∴ ΔmixH = Negative
15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?
Ans. According to Raoult’s law
$$\frac{\text{P}\degree \text{A}-\text{Ps}}{\text{P}\degree \text{A}}=\frac{n_B}{n_A}=\frac{W_{B}}{M_{B}}×\frac{\text{M}_{A}}{\text{W}_{A}}$$
P°A (for water) = 1.013 bar; PS = 1.004 bar; WS = 2g; WA = 100 – 2 = 98g; MA = 18g mol–1.
$$\frac{(1.013-1.004)\text{bar}}{(1.013\space\text{bar})}=\frac{(2g)×(18 \text{g mol}^{\normalsize-1})}{\text{Mg}×(98 g)}\\\therefore\space\text{M}_{B}=\frac{(2g)×(18\text{g mol}^{\normalsize-1})×(1.013\space\text{bar})}{(0.009\text{bar})×(98 g)}=41.35\text{g mol}^{\normalsize-1}$$
Hence, the molecular mass of the solute is 41.35g mol–1.
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 26.0g of heptane and 35.0g of octane?
Ans. Molar mass of Heptane (C7H16) = 7 × 12 + 16 × 1 = 100g mol–1
Molar mass of Octane (C8H18) = 8 × 12 + 18 × 1 = 114g mol–1
$$\text{Moles of Heptane present in mixture =}\frac{26.0}{100}=0.26\space\text{mol}\\\text{Moles of Octane present in mixture =}\frac{35.0}{114}=0.307\space\text{mol}\\\text{Mole fraction of Heptane x}_H =\frac{0.26}{0.26+0.307}=0.458$$
Mole fraction of Octane, xO = (1 – 0.458) = 0.542
Vapour pressure of Heptane = xH × P°
= 0.458 × 105.2 kPa = 48.18kPa
Vapour pressure of Octane = xO × P° = 0.542 × 46.8 kPa = 25.36 kPa
Vapour pressure of mixture = 48.18 + 25.36 = 73.54 kPa.
Hence, the vapour pressure of a mixture is 73.54 kPa.
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.
Ans. 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.
$$\therefore\space\text{Moles of water =}\frac{1000}{8}=55.5\space\text{moles}\\\therefore\space\text{Mole fraction of solute =}\frac{1}{1+55.5}=0.0177\\\text{Now,}\space\frac{\text{P}\degree-\text{P}_\text{S}}{\text{P}\degree}=x_2\\\frac{12.3-\text{P}_S}{12.3}=0.0177$$
⇒ Ps = 12.08 kPa
Hence, vapour pressure of 1 molal solution of a non-volatile solute in it is 12.08 kPa.
18. Calculate the mass of a non-volatile solute (molar mass 40g mol–1) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%.
Ans. According to Raoult’s Law,
$$\frac{\text{P}\degree_\text{A}-\text{P}\degree_\text{S}}{\text{P}\degree_\text{A}}=\frac{n_{B}}{n_{A}}=\frac{\text{W}_{B}}{\text{M}_{B}}×\frac{\text{M}_{A}}{\text{W}_{A}}$$
Let P°A = 1 atm, PS = 0.8 atm; P°A – PS = 0.2 atm; MB = 40g mol–1; WA = 114g; MA (C8H18) = 114g mol–1.
$$\text{W}_{B}=\frac{(\text{P}\degree_\text{A}-\text{P}_\text{S})}{\text{P}_\text{S}}×\frac{\text{M}_\text{B}×\text{W}_\text{A}}{\text{M}_\text{A}}\\=\frac{(0.2\space\text{atm})}{(1.0\space\text{atm})}×\frac{(40g\space\text{mol}^{\normalsize-1})×(114g)}{(114\text{g mol}^{\normalsize-1})}$$
= 8.0g
Hence, the mass of a non-volatile solute is 8.0g.
19. A solution containing 30g of non-volatile solute exactly in 90g 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.
Ans. (i) Let the molar mass of solute = Mg mol–1
$$\therefore\space\text{Moles of solute present =}\frac{30g}{\text{Mg mol}^{\normalsize-1}}=\frac{30}{\text{M}}\space\text{mol}\\\text{Moles of solvent present; (n}_1) =\frac{90}{18}=5\space\text{moles}\\\therefore\frac{\text{P}\degree-\text{P}_\text{S}}{\text{P}\degree}=\frac{n_2}{n_1+n_2}\\\frac{\text{P}\degree-2.8}{\text{P}\degree}=\frac{30/\text{M}}{5+30/\text{M}}\\1-\frac{2.8}{\text{P}\degree}=\frac{30}{(5\text{M}+30)}\\1-\frac{30}{5\text{M}+30}=\frac{2.8}{\text{P}\degree}\\1-\frac{6}{\text{M}+6}=\frac{2.8}{\text{P}\degree}$$
$$\frac{\text{M}+6-6}{\text{M}+6}=\frac{2.8}{\text{P}\degree}\\\frac{\text{M}}{\text{M}+6}=\frac{2.8}{\text{P}\degree}\\\frac{\text{P}\degree}{2.8}=1+\frac{6}{\text{M}}\space...\text{(i)}\\\text{After adding 18g of water,}\\\text{Moles of water becomes =}\frac{90+18}{18}=\frac{108}{18}=6\space\text{moles}\\\therefore\space\frac{\text{P}\degree-\text{P}_\text{S}}{\text{P}\degree}=\frac{30/\text{M}}{6+30/\text{M}}\\\text{P}_S \text{New vapour pressure = 2.9 kPa}\\\frac{\text{P}\degree-2.9}{\text{P}\degree}=\frac{30/\text{M}}{\text{M}(6\text{M}+30)}=\frac{5}{\text{M}+5}\\1-\frac{2.9}{\text{P}\degree}=\frac{5}{\text{M}+5}$$
$$1-\frac{5}{\text{M}+5}=\frac{2.9}{\text{P}\degree}\\\frac{\text{M+5-5}}{\text{M+5}}=\frac{2.9}{\text{P}\degree}\\\frac{\text{P}\degree}{2.9}=\frac{\text{M}+5}{\text{M}}\Rarr=1+\frac{5}{\text{M}}\\\frac{\text{P}\degree}{2.9}=1+\frac{5}{\text{M}}\space...\text{(ii)}\\\text{Dividing equation (i) by (ii), we get,}\\\frac{2.9}{2.8}=\frac{1+6/\text{M}}{1+5/\text{M}}\\2.9\bigg(1+\frac{5}{\text{M}}\bigg)=2.8\bigg(1+\frac{6}{\text{M}}\bigg)\\2.9+\frac{2.9×5}{\text{M}}=2.8+\frac{2.8×6}{\text{M}}$$
$$2.9+\frac{14.5}{\text{M}}=2.8+\frac{16.8}{\text{M}}\\0.1=\frac{16.8}{\text{M}}-\frac{14.5}{\text{M}}=\frac{2.3}{\text{M}}\\\text{M}=\frac{2.3}{0.1}$$
M = 23g mol–1
(ii) Putting M = 23, in equation (i), we get
$$\frac{\text{P}\degree}{2.8}=1+\frac{6}{23}=\frac{29}{23}\\\text{P\degree}=\frac{29}{23}×2.8=3.53\text{kPa.}$$
Hence, the molar mass of solute is 23g mol–1 and vapour pressure of water at 298 K is 3.53 kPa.
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.
Ans. Mass of sugar in 5% (by mass) solution means 5g in 100g of solvent (water)
Molar mass of sugar = 342g mol–1
$$\text{Molality of sugar solution =}\frac{5×1000}{342×100}=0.146\\\therefore\space\Delta\text{T}_f\text{for sugar solution = 273.15 – 271 = 2.15°}\\\Delta \text{T}_f=\text{K}_f×m\\\Delta\text{T}_f=\text{K}_f×0.146\\\Rarr\space\text{K}_f=\frac{2.15}{0.146}\\\text{Molality of glucose solution}=\frac{5}{180}×\frac{1000}{100}=0.278\space\text{(Molar mass of glucose = 180g mol}^{\normalsize–1})\\\Delta \text{T}_f=\text{K}_f×m=\frac{2.15}{0.146}×0.278=4.09\degree$$
∴ Freezing point of glucose solution = 273.15 – 4.09 = 269.06K.
Hence, the freezing point of glucose solution is 269.06 K.
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.
$$\textbf{Ans}\space\text{Using the relation,}\space\text{M}_{2}=\frac{1000×k_f×w_2}{w_1×\Delta\text{T}_f}\\\text{M}_{\text{AB}_2}=\frac{1000×5.1×1}{20×2.3}=110.87\text{mol}^{\normalsize-1}\\\text{M}_{\text{AB}_{4}}=\frac{1000×5.1×1}{20×1.3}=196.15\text{g mol}^{\normalsize-1}$$
Let the atomic masses of A and B are ‘p’ and ‘q’ respectively.
Then, molar mass of AB2 = p + 2q = 110.87g mol–1 ...(i)
And molar mass of AB4 = p + 4q = 196.15g mol–1 ...(ii)
Substracting equation (ii) from equation (i), we get
2q = 85.28
⇒ q = 42.64
Putting q = 42.64 in equation (i), we get
p = 110.87 – 85.28
p = 25.59
Hence, atomic mass of A = 25.59g mol–1 and atomic mass of B = 42.64g mol–1
22. At 300 K, 36g glucose present per litre in its solution has 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?
$$\textbf{Ans}.\space\pi=\text{CRT}=\frac{\text{W}_{B}×R×T}{\text{M}_{B}×V}\\\text{For both the solution R, T and V are constants}\\\textbf{I}^{\textbf{st}}\textbf{case:}\space(4.98\space\text{bar})=\frac{(36g)×R×T}{(180\text{g mol}^{\normalsize-1})×\text{V}}\\\text{(II)}^{\textbf{nd}}\textbf{case}:\space(1.52\space\text{bar})=\frac{\text{W}_{B}×\text{R}×\text{T}}{\text{M}_{\text{B}}×\text{V}}\\\text{Divide equation (ii) by equation (i)}\\\frac{(1.52 \text{bar})}{(4.98\text{bar})}=\frac{\text{W}_{B}}{\text{M}_{B}}×(5\text{mol})\\\text{or}\space\frac{\text{W}_{B}}{\text{M}_{B}}=\frac{1.52}{4.98}×\frac{1}{(5 \text{mol})}$$
= 0.610 mol–1
Hence, its concentration is 0.610 mol–1.
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) NaClO4and water
(iv) Methanol and acetone
(v) Acetonitrile (CH3CN) and acetone (C3H6O)
Ans. (i) Both n-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) NaClO4 is an ionic compound and gives Na+ and ClO4– 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 hydrogen bonding.
(v) Both CH3CN and C3H6O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
24. Based on solute solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Ans. (i) n-Octane is non polar and can dissolve non-polar solutes. It cannot dissolve polar (and ionic) solutes.
(ii) Cyclohexane is non-polar. Hence, easily soluble in n-octane.
(iii) Methanol and acetonitrile are polar and have very low solubility in n-octane.
(iv) KCl is ionic compound and hence, insoluble in n-octane.
The increasing order for solubility in n-octane is as follows:
KCl < CH3OH < CH3CN < Cyclohexane
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
Ans. (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.
26. If the density of lake water is 1·25 g mL–1, and it contains 92g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
$$\textbf{Ans.}\space\text{Molality of Na}^{+} \text{ions (m) }=\frac{\text{No. of mole of Na}^{+}\text{ions}}{\text{Mass of water in kg}}\\=\frac{(92g)/(23g\space \text{mol}^{\normalsize-1})}{1 \text{kg}}=4\text{mol kg}^{\normalsize-1}=4\text{m}$$
Hence, the molality of Na+ ions in the lake is 4m.
27. If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
Ans. CuS ⇌Cu2+ + S2–, Ksp = 6 × 10–16
Maximum molarity of CuS in aqueous solution means solubility of CuS.
Let the solubility of CuS be S mol L–1
28. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5g of CHO is dissolved in 450g of CH3CN.
Ans. Mass of aspirin = 6.5g
Mass of acetonitrile = 450g
Then, Total mass of the solution = (6.5 + 450)g = 456.5g
$$\text{Therefore, Mass percentage of C}_9\text{H}_8\text{O}_4=\frac{6.5}{456.5}×100\%$$
= 1.424%
Hence, the mass percentage is 1.424 %.
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.
Ans. 1.5 × 10–3 m aqueous solution of nalorphene means that 1.5 × 10–3 mole of nalorphene is dissolved in 1 kg of water.
Molar mass of nalorphene, C19H21NO3 = 19 × 12 + 21 × 1 + 1 × 14 + 3 × 16 = 311g mol–1
$$\text{Number of moles of solute =}\frac{\text{Mass}}{\text{Molar Mass}}=\frac{1.5×10^{3}g}{311\text{g mol}^{\normalsize-1}}=4.8×10^{\normalsize-6}\text{mol}\\\text{Molality =}\frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}}\\\text{Mass of solvent =}\frac{\text{Number of moles of solute}}{\text{Molality}}=\frac{4.8×10^{\normalsize-6}\text{mol}}{1.5×10^{\normalsize-3}\text{mol kg}}=0.0032\text{kg}$$
= 3.2g
Hence, the mass of 1.5 × 10–3 m aqueous solution is 3.2g.
30. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0·15 M solution in methanol.
Ans. M = 0.15 M = 0.15 mol L–1
Molar mass of solute = 6 × 12 + 5 × 1 + 1 × 12 + 2 × 16 + 1 × 1 = 122g mol–1
Volume of solution = 250 mL = 0.25 L
$$\text{Molarity (M) =}\frac{\text{Mass of solute/molar mass}}{\text{Volume of solution in liters}}\\(0.15 \text{mol L}^{\normalsize-1})=\frac{\text{Mass of solute}}{(122 \text{g mol}^{\normalsize-1})×(0.25\text{L})}$$
Mass of solute = (0.15 mol L–1) × (122g mol–1) × (0.25 L) = 4.575g
Hence, the amount of benzoic acid required is 4.575g.
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.
Ans. The depression in freezing point of a solute in water depends upon the number of particles or ions in its aqueous solution or its degree of dissociation (a). Three acids are arranged in the order of their increasing acidic strengths as follows:
CH3COOH < CCl3COOH < CF3COOH
Fluorine is more electronegative than chlorine. So, trifluoracetic acid is stronger than trichloroacetic acid which in turn stronger than acetic acid.
32. Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250g of water. Ka = 1.4 × 10–3 Kf = 1.86 K kg mol–1.
Ans. Mass of solute (CH3CH2CHClCOOH) = 10g
Molar mass of CH3CH2CHClCOOH = 4 × 12 + 7 × 1 + 1 × 35.5 + 2 × 16 = 48 + 7 + 35.5 + 32 = 122.5g mol–1
$$\text{Molality =}\frac{\text{Mass of solute/molar mass of solute}}{\text{Mass of solvent (kg)}}\\=\frac{10g/122.5\text{g mol}^{\normalsize-1}}{0.25g}=0.326\space\text{m}$$
Let a be the degree of dissociation of CH3CH2CHClCOOH then
CH3CH2CHClCOOH ⇌ CH3CH2CHCICOO– + H+
| Initial conc. | C mol–1 kg | 0 | 0 |
| At equilibrium | C(1 – α) | Cα | Cα |
$$\therefore\space\text{K}_a=\frac{C\alpha.C\alpha}{C(1-\alpha)}\\=\frac{C\alpha^{2}}{1-\alpha}\space[\because\text{Considering (1 –} \alpha) \text{= 1 for dilute solution}]\\\text{Now,}\space\text{K}_a=\frac{C\alpha^{2}}{1}\\\Rarr\space\text{K}_a=\text{C}\alpha^{2}\\\Rarr\space\alpha=\sqrt{\frac{\text{K}_a}{\text{C}}}\\=\sqrt{\frac{1.4×10^{\normalsize-3}}{0.326}}\space(\because \text{K}_a=1.4×10^{\normalsize-3})$$
= 0.0655
CH3CH2CHClCOOH ⇌ CH3CH2CHClCOO– + H+
| Initial no. of moles | 1 | 0 | 0 |
| At equilibrium | 1 – α | α | α |
Total no. of moles after dissociation = 1 – α + α + α
= 1 + α
Van’t Hoff factor
Total no. of moles after dissociation
$$\text{(i)}=\frac{\text{Total no. of moles after dissociation}}{\text{No. of moles before ddissociation}}\\\text{i}=\frac{1+\alpha}{1}$$
= 1 + α
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
ΔTf = i.Kf m
= 1.0655 × 1.86 kg mol–1 × 0.326 mol kg–1
= 0.65 K
Hence, the depression in the freezing point of water is 0.65 K.
33. 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid.
Ans. Calculation of Van’t Hoff factor(i)
Given, w1 = 500g = 0.5 kg, w2 = 19.5g, Kf = 1.86 K kg mol–1, ΔTf = 1K
Molar mass of CH2FCOOH (M2) = 2 × 12 + 3 × 1 + 1 × 19 + 2 × 16
= 24 + 3 + 19 + 32
= 78 g mol–1
ΔTf= iKf m
$$i=\frac{\Delta T_f}{\text{K}_fm}\space...\text{(i)}\\\text{m}=\frac{w_2}{\text{M}_2×w_1}=\frac{19.5 g}{(78 \text{g mol}^{\normalsize-1})×(0.5\text{kg})}\\\text{= 0.5 mol kg}^{\normalsize–1}\space...\text{(ii)}\\\text{From equation (i), we get}\\ i=\frac{1}{(1.86\text{K kg mol}^{\normalsize-1})×(0.5\text{mol kg}^{\normalsize-1})}$$
= 1.0753
Calculation of dissociation constant, Ka
Let α be the degree of dissociation of CH2FCOOH then
CH2FCOOH ⇌ CH2FCOO– + H+
| Initial conc. | C mol L–1 | 0 | 0 |
| At equilibrium | C(1 – α) | Cα | Cα |
$$\text{Total = C}(1 + \alpha)\\\therefore\space i=\frac{\text{C}(1+a)}{\text{C}}\\\Rarr i=1+\alpha\\\Rarr\space\alpha=i-1\\=1.0753-1\\=0.0753\\\text{Now,}\space\text{K}_a=\frac{[\text{CH}_2\text{FCOO}^{\normalsize-}][\text{H}^{+}]}{[\text{CH}_2\text{FCOOH}]}\\=\frac{C\alpha.C\alpha}{\text{C}(1-\alpha)}=\frac{C\alpha^{2}}{1-\alpha}\\\text{Taking the volume of the solution as 500 mL, we have the concentration:}\\\text{C}=\frac{\frac{19.5}{78}}{500}×1000\text{M}$$
= 0.5 M
$$\text{Therefore,}\space\text{K}_a=\frac{C\alpha^{2}}{1-\alpha}=\frac{0.5×(0.0753)^{2}}{1-0.0753}=\frac{0.5×0.00567}{0.9247}$$
= 0.00307
= 3.07 × 10–3
Hence, the Van’t Hoff factor and dissociation constant of fluoroacetic acid 1.0753 and 3.07 × 10–3 respectively.
34. Vapour pressure of water at 293 K is 17·535 mm Hg. Calculate the vapour pressure of water at 293 K when 25g of glucose is dissolved in 450g of water.
Ans. According to Raoult’s Law
$$x_B=\frac{n_{B}}{n_{A}+n_{B}}\\n_{B}=\frac{W_{B}}{M_{B}}=\frac{25}{180}=0.1389\\\text{n}_A=\frac{W_{A}}{M_{A}}=\frac{450}{18}=25\\\frac{17.535-P_S}{17.553}=\frac{0.1389}{25+0.1389}$$
PS = 17.438 mm of Hg
Hence, the vapour pressure of water is 17.438 mm of Hg.
35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Ans. Here, p = 760 mm Hg, KH = 4.27 × 105 mm Hg (at 298 K)
According to Henry’s law, p = KHX
$$\text{X}=\frac{p}{\text{K}_\text{H}}\\=\frac{760\space\text{mm Hg}}{4.27×10^{5}\text{mm Hg}}$$
= 177.99 × 10–5
= 178 × 10–5
Hence, the mole fraction of methane in benzene is 178 × 10–5.
36. 100g of liquid A (molar mass 140g mol–1) was dissolved in 1000g of liquid B (molar mass 180g 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.
$$\textbf{Ans.}\space\text{Number of moles of liquid A,}\space n_{A}=\frac{\text{W}_{\text{A}}}{\text{M}_{\text{A}}}=\frac{100}{140}=0.714\space\text{mol}\\\text{Number of moles of liquid B,}\space n_\text{B}=\frac{\text{W}_\text{B}}{\text{M}_\text{B}}=\frac{1000}{180}=5.55\space\text{mol}\\\text{Then, mole fraction of A,}\space\text{X}_\text{A}=\frac{n_{\text{B}}}{n_{\text{A}}+n_{\text{B}}}=\frac{0.714}{0.714+5.55}=0.114$$
Mole fraction of B,
XB = 1 – 0.114 = 0.886
Vapour pressure of pure liquid B, P°B = 500 torr
Therefore, vapour pressure of liquid B in the solution,
PB = P°BXB
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution,
Ptotal = 475 torr = Ptotal = PA + PB
∴ Vapour pressure of liquid A in the solution,
PA = Ptotal – PB
= 475 – 443 = 32 torr
Now , PA = P°AXA
= PAXA = 320.114
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
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 Ptotal, Pcholroform and Pacetone as a function of Xacetone. The experimental data observed for different compositions of mixtures is:
| 100 × Xacetone | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| Pacetone/mm Hg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| Pchloroform/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.
Ans.
| 100 × Xacetone | 0 | 0.118 | 0.234 | 0.360 | 0.508 | 0.582 | 0.645 | 0.721 |
| Pacetone/mm Hg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| Pchloroform/mm Hg | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
| PTotal (mm Hg) | 632.8 | 603.1 | 579.7 | 562.4 | 580.9 | 600.0 | 615.9 | 642.5 |
Since, the plot Ptotal dips downwards the solution show negative deviation from Raoult’s law.
38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.
Ans. Molar mass of benzene (C6H6) = 6 × 12 + 6 × 1 = 78g mol–1
Molar mass of toluene (C6H5CH3 ) = 7 × 12 + 8 × 1 = 92g mol–1
$$\text{No. of moles present in 80g of benzene =}\frac{80}{78}\space\text{mol}=1.026\space\text{mol}\\\text{No. of moles present in 100g of toluene =}\frac{100}{92}\text{mol}=1.087\space\text{mol}\\\text{Mole fraction of benzene, XC}_6\text{H}_6, =\frac{1.026}{1.026+1.087}=0.486$$
∴ Mole fraction of toluene, XC6H5CH3 = 1 – 0.486 = 0.514
It is given that vapour pressure of pure benzene, P°C6H6 = 50.71 mm Hg
Vapour pressure of pure toluene, P°C6H5CH3 = 32.06 mm Hg
Therefore, partial vapour pressure of benzene,
Ptotal = XC6H6 × P°C6H6
= 0.486 × 50.71
= 24.645 mm Hg
Partial vapour pressure of toluene, PC6H5CH3 = XC6H5CH3 × P°C6H5CH3
= 0.514 × 32.06
= 16.479 mm Hg
Total vapour pressure of solution (p) = 24.645 + 16.479
= 41.124 mm Hg
$$\text{Mole fraction of benzene in vapour phase =}\frac{X_{C_6H_6}×\text{P}\degree C_6H_6}{\text{P}_{\text{total}}}=\frac{0.486×50.71}{41.124}$$
= 0.599 ≅ 0.6
Hence, the mole fraction of benzene in vapour phase 0.6.
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 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
Ans. Percentage of oxygen (O2) in air = 20%
Percentage of nitrogen (N2) 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 = 7600 mm
Therefore, partial pressure of oxygen,
$$\text{P}_{\text{O}_2}=\frac{20}{100}×7600\\\text{= 1520 mm Hg}\\\text{Partial pressure of nitrogen, PN}_2=\frac{79}{100}×7600\\\text{= 6004 mm Hg}\\\text{Now , according to Henry’s law,}\\\text{p = K}_H.X\\\text{For oxygen:}\\\text{P}_{\text{O}_2} = \text{K}_\text{H}.X_{\text{O}_2}\\\Rarr\space\text{X}_{\text{O}_2}=\frac{\text{P}_{\text{O}_2}}{\text{K}_H}\\=\frac{1520\space\text{mm Hg}}{3.30×10^{7}\text{mm Hg}}=4.61×10^{\normalsize-5}(\text{Given K}_H=3.30×10^{7}\text{mm})$$
For nitrogen:
PN2 = KH.XN2
$$\Rarr\space\text{X}_{N_2}=\frac{\text{P}_{\text{N}_2}}{\text{K}_H}=\frac{6004}{6.51×10^{7}}$$
= 9.22 × 10–5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10–5 and 9.22 × 10–5 respectively.
40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Ans. We know that, π = iCRT
$$\Rarr\space\pi= iCRT=\frac{in_B}{\text{V}}\text{RT}$$
Given, π = 0.75 atm
V = 2.5L
i = 2.47
T = (27 = 273)K = 300K
R = 0.0821L atm K–1 mol–1
$$\pi=\frac{in_B\text{RT}}{\text{V}}\\0.75=\frac{2.47×n_B×0.0821×300}{2.5}\\n_B=0.31\space\text{mol}\\\text{Molar mass of CaCl}_2 \text{(M}_B) = 1 × 40 + 2 × 35.5 = 111\text{g mol}^{\normalsize–1}\\n_B=\frac{W_B}{M_B}$$
WB = nB × MB = 0.031 × 111
= 3.42g
Hence, the required amount of CaCl2 is 3.42g
41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Ans. Step I: Calculate of Van’t Hoff factor (i) K2SO4 dissociated in water as:
$$\text{K}_2\text{SO}_4\xrightarrow{\text{(aq)}}\space 2\text{K}^{+}\text{(aq)}+\text{SO}_{4}^{2-}\text{(aq);}\alpha=\frac{i-1}{n-1}\\\alpha\text{(for complete dissociation) = 1, n = 3;}1=\frac{i-1}{3-1}\text{or i = 2 + 1 = 3}\\\textbf{Step II:}\space\text{Calculation of osmotic pressure} (\pi)\\\text{Osmotic pressure} (\pi) = i\text{CRT} =\frac{iW_B\text{RT}}{\text{M}_B}\\$$
i = 3; WB = 25 mg = 0.025g; MB = 2 × 39 + 32 + 4 × 16 = 174g mol<su[>–1; V = 2L; R = 0.0821 L atm K–1 mol–1
$$\pi=\frac{(2)×(0.025 g)×(0.0821\text{L atm K}^{\normalsize-1}\text{mol}^{\normalsize-1})×(298\text{K})}{(174 \text{g mol}^{\normalsize-1})×(2\text{L})}$$
= 5.27 × 10–3 atm.
Hence, the osmotic pressure of a solution is 5.27 × 10–3 atm.
Share page on
