Hello Students. Are you Searching for NCERT Solutions for Class 12 Chemistry Chapter 2? If yes then you are most welcome to NCERTian. Here we have provided you with the complete Question and Answers of Chapter 2: Solutions. These solutions are written by expert teachers and faculties keeping the need of students in mind.
| Chapter | 2. Solutions |
| Subject | Chemistry |
| Textbook | Chemistry I |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 2: Solutions.
NCERT Solutions for Class 12 Chemistry Chapter 2
Solutions Solution
Q1) Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer) A homogeneous mixture of two or more chemically non-reacting substances is called a solution. There are nine types of solution under the main 3 heads i.e., gaseous solution, liquid solution and solid solution.
There are three types of solutions:
- Gaseous solution: The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.
- Liquid solution: The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution.
- Solid solution: The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.
Q2) Give an example of a solid solution in which the solute is a gas.
Answer) In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.
Q3) Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer)
- Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.
- Molality: Molality (m) is defined as the number of moles of the solute per kilogram of the solvent.
- Molarity: Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution.
- Mass percentage: The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution.
Q4) Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL−1?
Answer) Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1
Then, number of moles of HNO3 = 68/63 mol
= 1.079 mol
Given,
Density of solution = 1.504 g mL−1
(Mass / Density) = Volume of 100 g solution = 100/1.504mL
= 66.49mL
= 66.49×10-3L
Molarity of solution = 1.079/66.49×10-3L
= 16.23M
Q5) A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL−1, then what shall be the molarity of the solution?
Answer) 10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 − 10) g = 90 g of water.
Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1
Then, number of moles of glucose = 10/180mol
= 0.056 mol
⇒ Molality of solution = 0.056mol/0.09kg = 0.62m
Number of moles of water = 90g/18g mol-1
= 5 mol
⇒ Mole fraction of glucose (xg) = 0.056/0.056 + 5
= 0.011
And, mole fraction of water = 0.056mol/83.33×10-3L
= 1 – 0.011
= 0.989
If the density of the solution is 1.2 g mL−1, then the volume of the 100 g solution can be given as:
= 100g/1.2gmL-1
= 83.33mL
83.33×10-3 L
⇒ Molarity of the solution = 0.056mol/83.33×10-3L
= 0.67 M.
Q6) 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?
Answer) Let the amount of Na2CO3 in the mixture be x g.
Then, the amount of NaHCO3 in the mixture is (1 − x) g.
Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16
= 106 g mol−1
⇒ Number of moles Na2CO3 = x/106mol
Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16
= 84 g mol−1
⇒ Number of moles of NaHCO3 = (1-x)/84 mol
According to the question,
x/106 = (1-x)/84
⇒ 84x = 106 − 106x
⇒ 190x = 106
⇒ x = 0.5579
Therefore, number of moles of Na2CO3 = 0.5579/106 mol
= 0.0053 mol
And, number of moles of NaHCO3 = (1-0.5579)/84
= 0.0053 mol
HCl reacts with Na2CO3 and NaHCO3 according to the following equation
2HCL + Na2CO3 → 2NaCl + H2O + CO2
HCL + NaHCO3 → NaCl + H2O + CO2
1 mol of Na2CO3 reacts with 2 mol of HCl.
Therefore, 0.0053 mol of Na2CO3 reacts with 2 × 0.0053 mol = 0.0106 mol.
Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl.
Therefore, 0.0053 mol of NaHCO3 reacts with 0.0053 mol of HCl.
Total moles of HCl required = (0.0106 + 0.0053) mol
= 0.0159 mol
In 0.1 M of HCl,
0.1 mol of HCl is preset in 1000 mL of the solution.
Therefore, 0.0159 mol of HCl is present in (1000×0.0159)/0.1 mol
= 159 mL of the solution
Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both.
Q7) A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer) Total amount of solute present in the mixture is given by,
300 x (25/100) + 400 x (40/100)
= 75 + 160
= 235 g
Total amount of solution = 300 + 400 = 700 g
Therefore, mass percentage (w/w) of the solute in the resulting solution,
= 235/700 x 100%
= 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution,
= (100 − 33.57)%
= 66.43%
Q8) An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL−1, then what shall be the molarity of the solution?
Answer) Molar mass of ethylene glycol [C2H4(OH)2] = 2 × 12 + 6 × 1 + 2 ×16
= 62 gmol−1
Number of moles of ethylene glycol = 222.6g/62gmol-1
= 3.59 mol
Therefore, molality of the solution = 3.59 mol/0.200kg
= 17.95 m
Total mass of the solution = (222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 g mL−1
⇒ Volume of the solution = 422.6 g/1.072g mL-1
= 394.22 mL
= 0.3942 × 10−3 L
Therefore, Molarity of the solution = 3.59 mol/(0.39422×10-3L)
= 9.11 M
Q9) 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.
Answer i) 5 ppm (by mass) means 15 parts per million (106) of the solution.
Therefore, percent by mass = (15/106)x100%
= 1.5 × 10−3 %
Answer ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5 g mol−1
Now, according to the question,
15 g of chloroform is present in 106 g of the solution.
i.e., 15 g of chloroform is present in (106 − 15) ≈ 106 g of water.
Molality of the solution = (15/119.5 mol) x 1/(106 x 10-3 kg)
= 1.26 × 10−4 m.
Q10) What role does the molecular interaction play in a solution of alcohol and water?
Answer) In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.
Q11) Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer) When gases dissolve in water, the process is accompanied by release of excess heat energy, i.e., exothermic. According to Le Chatelier’s principle, when the temperature of the process is increased further the equilibrium shifts in backward direction. Hence, gases become less soluble in liquids.
Q12) State Henry’s law and mention some important applications?
Answer) Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as: p = KH x
Where, KH is Henry’s law constant
Applications of Henry’s law:
- In the sea diving
- In the production of carbonated beverages
Q13) The partial pressure of ethane over a solution containing 6.56 × 10−3 g of ethane is 1 bar. If the solution contains 5.00 × 10−2 g of ethane, then what shall be the partial pressure of the gas?
Answer) Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1
= 30 g mol−1
Therefore, Number of moles present in 6.56 × 10−3 g of ethane = (6.56×10-3)/30
= 2.187 × 10−4 mol
Let the number of moles of the solvent be x.
According to Henry’s law,
p = KHx
⇒ 1 bar = KH x (2.187×10-4) / (2.187×10-4 + x)
⇒ 1 bar = (KH x 2.187×10-4)/x (since, x >> 2.187×10-4)
⇒ KH = (x/2.187×10-4)bar
Number of moles present in 5.00 × 10−2g of ethane = (5.00×10-2)/30 mol
= 1.67 × 10−3 mol
According to Henry’s law,
p = KHx
(5.00×10-2)/30 mol
= 7.636 bar
Hence, partial pressure of the gas shall be 7.636 bar.
Q14) What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔsolH related to positive and negative deviations from Raoult’s law?
Answer) Positive deviation from Raoult’s law:
Solutions exhibit positive deviation from Raoult’s law when they have vapor pressure more than expected from the law.
Here, ΔmixH is positive as the energy is consumed for breaking the strong interaction and forming weaker interactions. In a similar way, ΔmixV is positive as the expansion of volume takes place.
Negative Deviation from Raoult’s Law:
Solutions exhibit negative deviation from Raoult’s law when they have vapor pressure less than expected from the law.
Here, ΔmixH is negative as the energy is released due to replacement of weaker interactions by stronger ones.
Q15) 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) Given that,
Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar
Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar
Mass of solute, (w2) = 2 g
Mass of solvent (water), (w1) = 98 g
Molar mass of solvent (water), (M1) = 18 g mol−1
According to Raoult’s law,
Hence, the molar mass of the solute is 41.35 g mol−1.
Q16) Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer) Vapour pressure of heptane (p10) = 105.2kPa
Vapour pressure of octane (p20) = 46.8 kPa
We know that,
Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1
= 100 g mol−1
Therefore, Number of moles of heptane = 26/100 mol
= 0.26 mol
Molar mass of octane (C8H18) = 8 × 12 + 18 × 1
= 114 g mol−1
Therefore, Number of moles of octane = 35 /114 mol
= 0.31 mol
Mole fraction of heptane, x1 = 0.26/(0.26 + 0.31)
= 0.456
And, mole fraction of octane, x2 = 1 − 0.456
= 0.544
Now, partial pressure of heptane,
p1 = x1p10
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane,
p2 = x2p20
= 0.544 × 46.8
Hence, vapour pressure of solution, ptotal = p1 + p2
= 47.97 + 25.46
= 73.43 kPa
Q17) 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) We have given 1 molal solution i.e., 1 mole of non-volatile solute in 1000 g water
Vapor pressure of water, P0P0 = 12.3 kPa
Now,
Molar mass of water = 18 g/mol
Number of moles of water in the solution = 1000/18 = 55.56mol
Mole fraction of solute = 1/(1 + 55.56) = 0.0176
Thus,
By Raoult’s law,
(P0 – PS)/P0 = x
(12.3 – Ps)/12.3 = 0.0176
Thus,
V. P. of solution, Ps = 12.082kPa
Q18) Calculate the mass of a non-volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g octane to reduce its vapor pressure to 80%.
Answer) Let the vapour pressure of pure octane be
Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80/100p10 = 0.8p10
Molar mass of solute, M2 = 40 g mol−1
Mass of octane, w1 = 114 g
Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1
= 114 g mol−1
Applying the relation,
Hence, the required mass of the solute is 8 g.
Q19) 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) vapor pressure of water at 298 K.
Answer i) Given that,
Mass of non-volatile solute = 30 g
Let molar mass of the same be M g/mol
Thus, number of moles of solute = 30/M mol
Case 1
Mass of water = 90 g
Molar mass of water = 18 g/mol
So, number of moles of water involved 90/18 = 5 mol
Vapor pressure of solution, Ps = 2.8 kPa
Mole fraction of solute = (30/M)/(30/M) + 5
By Raoult’s law,
(P0 – PS)/P0 = x
(P0 – 2.8)/P0 = (30/M)/(30/M) + 5
Therefore, P0/2.8 = 1 + 6/M
Case 2
Mass of water = 90 + 18 = 108 g
Molar mass of water = 18 g/mol
So, number of moles of water involved = 109/18 = 6mol
Vapor pressure of solution, Ps = 2.9 kPa
Mole fraction of solute = (30/M)/(30/M) + 6
By Raoult’s law,
(P0 – PS)/P0 = x
(P0 – 2.8)/P0 = (30/M)/(30/M) + 6
P0/2.9 = 1 + 5/M
Now,
Considering both the above equations, we get-
2.9/2.8 = (1 + 6/M)/(1 + 5/M)
Therefore, Molar mass of solute; M = 23 g/mol
Answer ii) vapor pressure of water at 298 K.
Now, putting this value in equation (1.1)
P0/2.8 = 1 + 6/M = 1 + 6/23
Therefore, V.P. of water, P0=3.53kPa.
Q20) 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.
Answer) Here, ΔTf = (273.15 − 271) K
= 2.15 K
Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 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
= 5/342mol
= 0.0146 mol
Therefore, molality of the solution,
m = 0.0146 mol/0.095 kg
= 0.1537 mol kg−1
Applying the relation,
ΔTf = Kf × m
⇒ Kf = ΔTf/m
= 2.15K/0.1537 mol kg -1
= 13.99 K kg mol−1
Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 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 = 5/180 mol
= 0.0278 mol
Therefore, molality of the solution, m = 0.0278 mol/0.095kg
= 0.2926 mol kg−1
Applying the relation,
ΔTf = Kf × m
= 13.99 K kg mol−1 × 0.2926 mol kg−1
= 4.09 K (approximately)
Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K.
Q21) 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 Kwhereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol−1. Calculate atomic masses of A and B.
Answer) We know that,
M2 = (1000 x w2 x kf)/ΔTf x w1
Then, MAB2 = (1000 x 1 x 5.1)/2.3 x 20
= 110.87 g mol−1
MAB4 = (1000 x 1 x 5.1)/(1.3 x 20)
= 196.15 g mol−1
Now, we have the molar masses of AB2 and AB4 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.
Q22) 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)
Here,
T = 300 K
π = 1.52 bar
R = 0.083 bar L K−1 mol−1
Applying the relation,
π = CRT
⇒ C = π/RT
= 1.52 bar / (0.083 bar L K-1 mol-1 x 300K)
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
Q23) 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 i) Both are nonpolar and hence, the intermolecular interactions will be London dispersion forces.
Answer ii) Both are nonpolar and hence, the intermolecular interactions will be London dispersion forces.
Answer iii) The intermolecular interactions will be ion-dipole interactions as NaClO4 is an ionic compound and water is a polar molecule.
Answer iv) Both are polar molecules and hence, intermolecular interactions will be dipole-dipole interactions.
Answer v) Both are polar molecules and hence, intermolecular interactions will be dipole-dipole interactions.
Q24) Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer) 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
Q25) 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)
Answer i) Phenol (C6H5OH) has the polar group −OH and non-polar group −C6H5. Thus, phenol is partially soluble in water.
Answer ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water.
Answer iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water. Thus, formic acid is highly soluble in water.
Answer iv) Ethylene glycol has polar −OH group and can form H−bond. Thus, it is highly soluble in water.
Answer v) Chloroform is insoluble in water.
Answer vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky non-polar −C5H11 group. Thus, pentanol is partially soluble in water.
Q26) 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) Number of moles present in 92 g of Na+ ions = 92g/23 g mol-1
= 4 mol
Therefore, molality of Na+ ions in the lake = 4 mol / 1 kg
= 4m
Q27) If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of CuS in aqueous solution.
Answer) Solubility product of CuS, Ksp = 6 × 10−16
Let s be the solubility of CuS in mol L−1.
CuS ↔ [Cu2+] + [S2−]
= s × s
= s2
Then, we have, Ksp = s2 = 6×10−16
s = 2.45 × 10−8 mol L−1
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10−8 mol L−1.
Q28) Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Answer) 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 ofC9H8O4 = (6.5/456.5) x 100%
= 1.424%
Q29) 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) 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 × 10−3m aqueous solution of nalorphene,
1 kg (1000 g) of water contains 1.5 × 10−3 mol = 1.5 x 10-3 x 311g
= 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, mass of the solution containing 1.5 mg of nalorphene is:
(1000.4665 × 1.5 × 10−3)/0.4665 g
= 3.22g
Hence, the mass of aqueous solution required is 3.22 g.
Q30) Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer) 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 x 250)/1000 mol of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16
= 122 g mol−1
Hence, required benzoic acid = 0.0375 mol × 122 g mol−1
= 4.575 g
Q31) 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) 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.
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)
Q33) 19.5 g of CH2FCOOH is dissolved in 500 g 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.
Answer)
Q34) 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)
Q35) Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.
Answer)
Q36) 100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer)
Q37) 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 ptotal’ pchloroform’ and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture 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 |
Answer) From the question, we have the following data
| 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 |
It can be observed from the graph that the plot for the ptotal of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.
Q38) 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 80 g of benzene is mixed with 100 g of toluene.
Answer)
Q39) 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 the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
Answer)
Q40) 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.
Answer)
Q41) Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 liter of water at 25° C, assuming that it is completely dissociated.
Answer)
That’s it. These were the solutions of NCERT Class 12 Chemistry Chapter 2 – Solutions. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.