In this post, questions and answers of class 12 chapter Solutions are provided which included NCERT questions.
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.
Answer: 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. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Answer: 30% by mass of C6H6 in CCl4 = 30 g C6H6 in 100 g solution
Mass of C6H6 = 30 g
Mass of CCl4 = Mass of solution – Mass of C6H6 = 100 – 30 = 70 g
No. of moles of C6H6 = 30/78 = 0.385
No. of moles of CCl4 = 70/154 = 0.455
So, the mole fraction of benzene is 0.458
3. Calculate the molarity of each of the following solution:
(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
Answer: (a) Mass of Co(NO3)2.6H2O = 30 g
Molar mass of Co(NO3)2.6H2O = 59 + 2(14+3(16)) + 6(2(1)+16) = 291 g mol-1
Number of moles of Co(NO3)2.6H2O = Given mass/Molar mass = 30/291 = 0.1031 mol
Volume of solution = 4.3 L
Molarity = Number of moles/Volume of solution = 0.1031/4.3 = 0.02397 M = 0.024 M
(b) Molarity of undiluted H2SO4 (M1) = 0.5 M
Volume of undiluted H2SO4 (V1) = 30 ml
Volume of diluted H2SO4 solution (V2) = 500 ml
Molarity of diluted solution (M2) is calculated as:
M1V1=M2V2
M2 = M1V1/V2 = (0.5 x 30)/(500) = 0.03 M
4. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Answer: Molality = 0.25 m
Mass of solvent = 2.5 Kg
Number of mole of urea = Molality x Mass of solvent (in Kg) = 0.25 x 2.5 = 0.625 mol
Molar mass of urea = 60 g/mol
Mass of urea = No. of mole x Molar mass = 0.625 x 60 = 37.5 g
5. Calculate:
(a) molality
(b) molarity and
(c) mole fraction of KI,
if the density of 20% (m/m) aqueous KI solution is 1·202 g mL-1.
Answer: Mass of KI in 100 g of the solution = 20 g
Mass of solvent (water) = 100 – 20 = 80 g = 0.080 kg
Mass of solution = 100 g
Density of solution = 1.202 g/ml
Volume of solution = Mass/Density = 100/1.202 = 83.195 ml = 0.0832 L
Molar mass of KI = 39 + 127 = 166 g mol-1
Molar mass of H2O = 18 g mol-1
No. of mole of KI = Given mass/Molar mass = 20/166 = 0.1204 mol
No. of mole of water = 80/18 = 4.44 mol
(a) Molality = No. of mole of KI/Mass of solvent (in Kg) = 0.1204/0.080 = 1.505 m
(b) Molarity = No. of mole of KI/Volume of solution (in L) = 0.1204/0.0832 = 1.45 M
(c) Mole Fraction of KI = Mole of KI/Mole of (KI+Water) = 0.1204/(0.1204+4.44) = 0.0264
6. H2S, 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.
Answer: Solubility of H2S gas = 0.195 m = 0.195 mole in 1000 g of water
Mole of H2S = 0.195 mol
Mole of water = 1000/18 = 55.55 mol
Mole fraction of H2S, X(H2S) = Mole of H2S/Mole of (H2S + H2O) = 0.195/(0.195+55.55) = 0.0035
We already know that, Pressure at STP = 0.987 bar
Applying Henry’s law, P = KH * X(H2S)
KH = P/X(H2S) = 0.987/0.0035 = 282 bar
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.
Answer: KH = 1.67 x 108 Pa
P(CO2) = 2.5 atm = 2.5 x 1.01325 x 105 Pa
Applying Henry’s law,
Mole fraction of CO2, X(CO2) = P/KH = (2.5 x 1.01325 x 105)/(1.67 x 108) = 1.517 x 10-3
For 500 ml of soda water, mole of H2O = 500/18 =27.78 mol
Mole fraction of CO2 = Mole of CO2/Mole of H2O
Mole of CO2 = 1.517 x 10-3 x 27.78 = 42.14 x 10-3 mol
Mass of CO2 = 42.14 x 10-3 x 44 = 1.854 g
8. The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the vapour phase.
Answer: Vapour pressure of pure liquid A (P∘A) = 450 mm
Vapour pressure of pure liquid B (P∘B) = 700 mm
Total vapour pressure of the solution (P) = 600 mm
According to Raoult’s law, P = P∘AXA + P∘BXB = P∘AXA + P∘B(1-XA)
600 = 450XA + 700(1-XA)
600 = 700 – 250XA
XA = 0.40
XB = 1 – XA = 1-0.40 = 0.60
Mole fraction of A = 0.40
Mole fraction of B = 0.60
PA = P∘AXA = 450 * 0.40 = 180 mm
PB = P∘BXB = 700 * 0.60 = 420 mm
PT = PA + PB = 180+420 = 600 mm
Mole fraction of A in vapour phase = PA/PT = 180/600 = 0.30
Mole fraction of B in vapour phase = PB/PT = 420/600 = 0.70
9. Vapour pressure of pure water at 298 K is 23.8 mm 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.
Answer: P∘(H2O) = 23.8 mm Hg
Mass of urea (w2) = 50 g
Mass of water (w1) = 850 g
Molar mass of urea and water are 60 g/mol and 18 g/mol respectively.
Mole of urea (n2) = 50/60 = 0.83 mol
Mole of water (n1) = 850/18 = 47.22 mol
Relative lowering of vapour pressure is 0.017
Now, put P∘(H2O) = 23.8 mm Hg in the relative lowering of vapour pressure expression, we get
Ps = 23.39 mm Hg = 23.4 mm Hg
Hence, the vapour pressure of solution is 23.4 mm Hg and its relative lowering is 0.017
10. Boiling point of water at 750 mm Hg is 96.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Answer: T∘b = 100∘C
Tb = 96.63∘C
∆Tb = T∘b – Tb = 100 – 96.63 = 3.37∘C = 3.37 K (Temperature difference is same in both units)
Mass of water (w1) = 500 g = 0.5 Kg
Molar mass of sugar (M2) = 342 g/mol
Kb = 0.52 K Kg mol-1
∆Tb = Kb * m
w2 = 1108.2 g = 1.11 Kg
Mass of sucrose to be added is 1.11 Kg.
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. (Kf for CH3COOH) = 3·9 K kg mol-1)
Answer: Depression in melting point (∆Tf) = 1.5∘C = 1.5 K
Mass of acetic acid (w1) = 75 g = 0.075 Kg
Molar mass of C6H8O6 (M2) = 176 g/mol
Kf = 3.9 K Kg mol-1
w2 = 5.08 g
Mass of ascorbic acid to be dissolved is 5.08 g
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.
Answer: Mass of polymer = 1.0 g
Volume of solution = 450 ml = 0.45 L
Molar mass of polymer = 185000 g/mol
Temperature = 37∘C = (37+273) K = 310 K
Mole of polymer = Mass/Molar mass = 1.0/185000 = 5.41 x 10-6 mol
R = 8.314 x 103 Pa L K-1 mol-1
Concentration, C = Mole/Volume = (5.41 x 10-6)/0.45 =12.02 x 10-6 M
Osmotic pressure, π = C x R x T
π = 12.02 x 10-6 x 8.314 x 103 x 310 = 30.979 Pa
Hence, the osmotic pressure is 30.979 Pa
13. Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer: A solution is a homogeneous mixture of two or more chemically non-reacting substances.
Types of solutions: There are nine types of solutions. Examples:
Gaseous solutions: (a) Gas in gas Air, mixture of O2 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 CO2 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)
14. Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be?
Answer: The solution likely to be formed is interstitial solid solution.
15. Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Answer: (i) Mole fraction: It is defined as the ratio of the number of moles of the solute(or solvent) to the total number of moles in the solution. If B is the number of moles of solute dissolved in A moles of solvent, then
(ii) Molality: It is defined as the number of moles of a solute present in 1000g (1kg) of a solvent.
Molality = Number of moles of solute/Mass of solvent (in Kg)
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.
Molarity = Number of moles of solute/Volume of solution (in L)
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.
Mass percentage = (Mass of solute/Mass of solution) x 100
16. 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 1·504 g mL-1 ?
Answer: Mass of HNO3 in solution = 68 g
Molar mass of HNO3 = 63 g mol-1
Mass of solution = 100 g
Density of solution = 1·504 g mL-1
Volume of solution = Mass of solution/Density of solution = 100/1.504 = 66.5 ml = 0.0665 L
Mole of HNO3 = Mass/Molar mass = 68/63 = 1.08 mol
Molarity of HNO3 = No. of mole/Volume of solution (L) = 1.08/0.0665 = 16.24 M
17. 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 glucose solution means 10 g glucose in 100 g of solution.
Mass of glucose = 10 g
Mass of solution = 100 g
Mass of water = 100 – 10 = 90 g = 0.090 Kg
Mole of glucose (n2) = 10/180 = 0.055 mol
Mole of water (n1) = 90/18 = 5 mol
Molality = Mole of glucose/Mass of water (Kg) = 0.055/0.090 = 0.617 m
Mole fraction of glucose = n2/(n1+n2) = 0.055/(5+0.055) = 0.011
Mole fraction of water = 1-Mole fraction of glucose = 1-0.011 = 0.989
Volume of solution = Mass/Density = 100/1.2 = 83.33 ml = 0.0833 L
Molarity = Moles/Volume (L) = 0.055/0.0833 = 0.66 M
18. 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 x g of Na2CO3 and (1-x) g NaHCO3 is present in the mixture.
Molar mass of Na2CO3 = 106 g/mol
Molar mass of NaHCO3 = 84 g/mol
No. of mole of Na2CO3 = x/106 mol
No. of mole of NaHCO3 = (1-x)/84 mol
As given that the mixture contains equimolar amounts of both Na2CO3 and NaHCO3, so
Mole of Na2CO3 = Mole of NaHCO3
On solving, x = 0.558 g
Mass of Na2CO3 and NaHCO3 are 0.558 g and 0.442 g respectively.
No. of mole of Na2CO3 = 0.558/106 = 0.00526 mol
No. of mole of NaHCO3 = 0.442/84 = 0.00526 mol
Now, the balanced chemical equation of Na2CO3 and NaHCO3 with HCl are:
Na2CO3 + 2 HCl → 2NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
From the balanced chemical equation, 1 mol Na2CO3 reacts with 2 mol HCl while 1 mol NaHCO3 reacts with 1 mol HCl.
So, 0.00526 mol Na2CO3 reacts with HCl = 2 x 0.00526 = 0.01052 mol
and 0.00526 mol NaHCO3 reacts with HCl = 0.00526 mol
Total mole of HCl required = 0.01052 + 0.00526 = 0.01578 mol
Molarity of HCl = 0.1 M
Volume of HCl required = Mole of HCl/Molarity of HCl = 0.01578/0.1 = 0.1578 L = 157.8 ml
19. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.
Answer: Mass of one component = 25% of 300 = 75 g
Mass of other component = 40% of 400 = 160 g
After mixing both the solution,
Total mass of solute = 75 + 160 = 235 g
Total mass of solution = 300 + 400 = 700 g
Mass % of solute in the final solution = (235/700) x 100 = 33.57 %
Mass % of solvent in the final solution = 100 – 33.57 = 66.43 %
20. An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C2 H6O2 ) 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: Mass of ethylene glycol = 222.6 g
Molar mass of ethylene glycol = 62 g/mol
No. of mole of ethylene glycol = 222.6/62 = 3.59 mol
Mass of solvent = 200 g = 0.2 Kg
Molality = Mole of ethylene glycol/Mass of solvent = 3.59/0.2 = 17.95 m
Mass of solution = Mass of (solute + solvent) = 222.6 + 200 = 422.6 g
Volume of solution = Mass/Density = 422.6/1.072 = 394.21 ml = 0.39421 L
Molarity = Mole of ethylene glycol/Volume of solution = 3.59/0.39421 = 9.11 M
21. 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: 15 ppm means 15 parts in million (106) by mass in the solution.
(i) Mass percent = (15*100)/106 = 15 x 10-4 %
(ii) As only 15 g CHCl3 is present in 106 g of solution. Mass of solvent = 106 g = 103 Kg
Molar mass of CHCl3 = 119.5 g/mol
No. of mole of CHCl3 = 15/119.5 = 0.126 mol
Molality = No. of mole of CHCl3/Mass of solvent (Kg) = 0.126/103 = 1.26 x 10-4 m
22. What role does the molecular interaction play in solution of alcohol in water?
Answer: In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.
23. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer: 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 Le-Chatelier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.
24. State Henry’s law and mention some of its important applications.
Answer: Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
or
The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution. p = KH.x
where KH is Henry’s law constant.
Applications of Henry’s law:
(i) 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 producing a hissing noise or with a fizz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium in order 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.
25. 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: We know that m = KH * P
Case 1: 6.56*10-3 = KH * 1 ….(i)
Case 2: 5.00*10-2 = KH * P …..(ii)
From (i) KH = 6.56*10-3
and From (ii) KH = 5.00*10-2/P
So, 6.56*10-3 = 5.00*10-2/P
On solving, P = 0.762 bar
Hence, the partial pressure of the gas is 0.762 bar.
Read more: Chapter 1 Solid State Q and A
26. According to Raoult’s law, what is meant by positive and negative deviation and how is the sign of ∆solH related to positive and negative deviations from Raoult’s law?
Answer: 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.
27. 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?
Answer: P° (for water) = 1.013 bar, Ps = 1.004 bar
2% non-volatile solute means 2 g solute in 100 g solution (or 98 g solvent)
Mass of solute (wB) = 2 g
Mass of water (wA) = 98 g
Molar mass of water (MA) = 18 g/mol
According to Raoult’s law,
On solving, MB = 41 g mol-1
Hence, the molar mass of the solute is 41 g mol-1.
28. 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?
Answer: Vapour pressure of pure heptane = 105.2 kPa
Vapour pressure of pure octane = 46.8 kPa
Molar mass of heptane (C7H14) = 100 g mol-1
Molar mass of octane (C8H16) = 114 g mol-1
Mass of heptane = 26 g
Mass of octane = 35 g
Mole of heptane = Mass/Molar mass = 26/100 = 0.26 mol
Mole of octane = Mass/Molar mass = 35/114 = 0.307 mol
Total mole = Mole of (heptane + octane) = 0.26 + 0.307 = 0.567 mol
Mole fraction of heptane = Mole of heptane/Total mole = 0.26/0.567 = 0.458
Mole fraction of octane = 1 – Mole fraction of heptane = 1 – 0.458 = 0.542
Vapour pressure of heptane = 105.2 x 0.458 = 48.18 kPa
Vapour pressure of octane = 46.8 x 0.542 = 25.36 kPa
Vapour pressure of the mixture = V.P of (heptane + octane) = 48.18 + 25.36 = 73.54 kPa
Hence, the vapour pressure of the mixture is 73.54 kPa
29. 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: 1 molal solution of a non-volatile solute means 1 mol of non-volatile solute in 1000 g of solvent (water).
No. of mole of solute = 1 mol
No. of mol of solvent = 1000/18 = 55.55 mol
Total mole = 1 + 55.55 = 56.55 mol
Mole fraction of non-volatile solute, X2 = Mole of solute/Total mole = 1/56.55 = 0.0177
Now, (P°-Ps)/P° = X2
(12.3-Ps)/12.3 = 0.0177
Ps = 12.08 kPa
Hence, the vapour pressure is 12.08 kPa.
30. Calculate the mass of a non-volatile solute (molecular mass 40 g mol-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%.
Answer: P° = 1 atm, wA = 114 g, MA = 114 g/mol MB = 40 g/mol, Ps = 0.8 atm (1 atm pressure reduced to 80%)
On solving, wB = 10 g
Hence, the mass of non-volatile solute is 10 g.
31. 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.
Answer:
32. 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: Case 1: 5% solution of cane sugar (molar mass: 342 g/mol) means 5 g cane sugar in 100 g solution (or 95 g solvent).
Tf = 271K and T°f = 273.15 K
∆Tf = T°f – Tf = 273.15 – 271 = 2.15 K
Mole of cane sugar = 5/342 = 0.015 mol
Mass of solvent = 95 g = 0.095 Kg
Molality of cane sugar, m = Moles of cane sugar/Mass of solvent = 0.015/0.095 = 0.158 m
Now, ∆Tf = Kf*m
Putting values, Kf = ∆Tf/m = 2.15/0.158 = 13.608 K Kg mol-1
Case 2: 5% solution of glucose (molar mass: 180 g/mol) means 5 g glucose in 100 g solution (or 95 g solvent).
Mole of glucose = 5/180 = 0.0278 mol
Mass of solvent = 95 g = 0.095 Kg
Molality of glucose, m = 0.0278/0.095 = 0.293 m
∆Tf = Kf*m = 13.608 * 0.293 = 3.99 K
∆Tf = T°f – Tf
3.99 = 273.15 – Tf
Tf = 269.16 K
Hence, the freezing point of 5% glucose solution is 269.16 K.
33. 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 (C3H6O)
Answer: (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) 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 dipole-dipole interactions.
(v) Both CH3CN and C3H6O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
34. 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 (C8H18) is a non-polar liquid and solubility is governed by the principle that like dissolve like. Keeping this in view, the increasing order of solubility of different solutes is:
KCl < CH3OH < CH3C=N < C6H12 (cyclohexane).
35. 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: (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.
36. 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.
Answer: Mass of solute (AB2) = 1 g
Mass of solute (AB4) = 1 g
Mass of solvent = 20 g = 0.020 Kg
∆Tf (AB2) = 2.3 K
∆Tf (AB4) = 1.3 K
Using the equation,
where, M2 is the molar mass of solute, w2 is mass of solute and w1 is mass of solvent in Kg.
So, molar mass of AB2 and AB4 are 110.87 g/mol and 196.15 g/mol respectively.
Let the atomic mass of A and B be p and q respectively.
Then, molar mass of AB2 = p+2q = 110.87 …..(i)
and molar mass of AB4 = p+4q = 196.15 …..(ii)
On solving the equations (i) and (ii), we get
p = 25.59 g/mol and q = 42.64 g/mol
Hence, the atomic mass of A and B are 25.59 g/mol and 42.64 g/mol respectively.
37. At 300 K, 36 g glucose present per litre in its solution has osmotic pressure of 4·98 bar. If the osmotic pressure of the solution is 1·52 bar at the same temperature, what would be its concentration?
Answer: Case 1: T = 300 K, w2 = 36 g, M2 = 180 g/mol, V = 1 L, π = 4.98 bar
Mole of glucose, n = 36/180 = 0.2 mol
π = CRT = (n/V)RT
4.98 = 0.2 RT
RT = 24.9
Case 2: T = 300 K (same as case 1), π = 1.52 bar
π = CRT
C = π/RT = 1.52/24.9 = 0.061 M
So, the concentration of the solution is 0.061 M.
38. If the density of lake water is 1·25 g mL-1, and it contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
Answer: Mass of Na+ ions= 92 g
Atomic mass of Na = 23 g/mol
Mass of solvent = 1 Kg
No. of mole of Na+ ions = 92/23 = 4 mol
Molality = No. of mole/Mass of solvent = 4/1 = 4 m
39. If the solubility product of CuS is 6 x 10-16, calculate the maximum molarity of CuS in aqueous solution.
Answer: CuS ⇌ Cu2+ + S2-
Let the solubility be s M
So, Ksp = [Cu2+][S2-]
6 x 10-16 = (s)(s)
s = √(6 x 10-16) = 2.45 x 10-8 M
Hence, the maximum molarity of CuS is 2.45 x 10-8 M.
40. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5g of C9H8O4 is dissolved in 450 g of CH3CN.
Answer: Mass of aspirin = 6.5 g
Mass of acetonitrile = 450 g
Mass of solution = 450 + 6.5 = 456.5 g
Mass percent = (Mass of aspirin/Mass of solution) x 100 = (6.5/456.5) x 100 = 1.424%
41. 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 x 10-3 m aqueous solution required for the above dose.
Answer: Mass of nalorphene = 1.5 mg = 1.5 x 10-3 g
Molar mass of nalorphene = 311 g/mol
Mole of nalorphene = Mass/Molar mass = (1.5 x 10-3)/311 = 4.823 x 10-6 mol
Molality of nalorphene = 1.5 x 10-3 m
Mass of solvent (Kg) = Mole of nalorphene/Molality of nalorphene = (4.823 x 10-6)/(1.5 x 10-3) = 0.0032153 Kg = 3.2153 g
Mass of solution = Mass of nalorphene + Mass of solvent = (1.5 x 10-3) + 3.2153 = 3.2168 g = 3.22 g
Hence, the mass of aqueous solution is 3.22 g.
42. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0· 15 M solution in methanol.
Answer: Molarity = 0.15 M
Volume of solution = 250 ml = 0.25 L
Mole of benzoic acid = Molarity x Volume of solution = 0.15 x 0.25 = 0.0375 mol
Molar mass of benzoic acid = 122 g/mol
Mass of benzoic acid = Mole x Molar mass = 0.0375 x 122 = 4.575 g
43. 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: Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, trifluoracetic acid is the strongest, trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group.
Thus, CF3COOH ionizes to the largest extent while CH3COOH ionizes to minimum extent in water. Greater the extent of ionization greater is the depression in freezing point. Hence, the order of depression in freezing point will be CH3COOH < CCl3COOH < CF3COOH.
44. Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250g of water. Ka = 1.4 x 10-3 and Kf = 1.86 K kg mol-1.
Answer: Mass of solute (CH3CH2CHClCOOH) = 10 g
Molar mass of CH3CH2CHClCOOH = 4 × 12 + 7 × 1 + 1 × 35.5 + 2 × 16 = 48 + 7 + 35.5 + 32 = 122.5 g mol-1
Mole of solute = Mass/Molar mass = 10/122.5 = 0.0816 mol
Mass of solvent = 250 g = 0.25 Kg
Molality = Mole of solute/Mass of solvent (Kg) = 0.0816/0.25 =0.326 m
Let α be the degree of dissociation of CH3CH2CHClCOOH, then
Total no. of moles after dissociation = 1 – α + α + α = 1 + α
Van’t Hoff factor, i = Total no. of moles after dissociation/No. of moles before dissociation
i = 1+α/1 = 1 + α = 1 + 0.0655 = 1.0655
Hence, the depression in the freezing point of water, ΔTf = i x Kf x m = 1.0655 × 1.86 kg mol-1 × 0.326 mol kg-1 = 0.65K
45. 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’s Hoff factor and dissociation constant of fluoroacetic acid.
Answer: w1 = 500 g = 0.5 kg, w2 = 19.5 g, Kf = 1.86 K kg mol-1, ΔTf = 1 K
Molar mass of CH2FCOOH (M2) = 24 + 3 + 19 + 32 = 78 g mol-1
ΔTf = i Kf m
46. 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: wB = 25 g, MB = 180 g/mol, wA = 450 g, MA= 18 g/mol P° = 17.535 mm Hg
47. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer: P = 760 mm Hg, KH = 4.27 × 105 mm Hg (at 298 K)
According to Henry’s law, p = KH * X
X = P/KH = 760/4.27×105 = 177.99 × 10-5 = 178 × 10-5
Hence, the mole fraction of methane in benzene is 178 × 10-5.
48. 100g of liquid A (molar mass 140 g 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.
Answer: Number of moles of liquid A, nA = Mass/Molar mass = 100/140 = 0.714 mol
Number of moles of liquid B, nB = Mass/Molar mass = 1000/180 mol = 5.556 mol
Total moles, (nA + nB) = 0.714 + 5.556 = 6.27 mol
Mole fraction of A, XA = Mole of A/Total moles = 0.714/6.27 = 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°B * XB = 500 * 0.886 = 443 torr
Total vapour pressure of the solution, Ptotal = 475 torr
So, Vapour pressure of liquid A in the solution, PA = Ptotal – PB = 475 – 443 = 32 torr
Now, PA = P°A * XA
P°A = PA/XA = 32/0.114 = 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
49. 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, Pchloroform 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/mmHg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| Pchloroform/mmHg | 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.
Answer:
| 100 * Xacetone | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| Pacetone/mmHg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| Pchloroform/mmHg | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
| Ptotal/mmHg | 632.8 | 603.1 | 579.7 | 562.4 | 580.9 | 600.0 | 615.9 | 642.5 |
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.
50. 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.
Answer: Molar mass of benzene (C6H6) = 78g mol-1
Molar mass of toluene (C6H5CH3 ) = 92 g mol-1
No. of moles present in 80 g of benzene = 80/78 mol = 1.026 mol
No. of moles present in 100 g of toluene = 100/92 mol = 1.087 mol
Total moles = Mole of (benzene + toluene) = 1.026 + 1.087 = 2.113 mol
Mole fraction of benzene, XC6H6, = Mole of benzene/Total mole = 1.026/2.113 = 0.486
Mole fraction of toluene, XC6H5CH3 = 1-XC6H6 = 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
So, Partial vapour pressure of benzene, PC6H6 = XC6H6 × P°(C6H6) = 0.486 × 50.71 = 24.645 mm Hg
and Partial vapour pressure of toluene, PC6H5CH3 = XC6H5CH3 × P°(C6H5CH3) = 0.514 × 32.06 = 16.479 mm Hg
Total vapour pressure of solution Ptotal = 24.645 + 16.479 = 41.124 mm Hg
Mole fraction of benzene in vapour phase = XC6H6×P°C6H6/Ptotal = 0.486×50.71/41.124 = 0.599
51. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with an 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 x 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in water.
Answer: 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, i.e., (10 × 760) mm = 7600 mm.
Therefore, partial pressure of oxygen, P(O2) = (20 x 7600)/100 = 1520 mm Hg
Partial pressure of nitrogen, P(N2) = (79 x 7600)/100 = 6004 mm Hg
Now, according to Henry’s law, P = KH.X
For oxygen, mole fraction of O2, X(O2) = P/KH= 1520/(3.30 x 107) = 4.61 x 10-5
For nitrogen, mole fraction of N2, X(N2) = P/KH= 6004/(6.51 x 107) = 9.22 x 10-5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10-5 and 9.22 × 10-5 respectively.
52. 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: π = 0.75 atm, V = 2.5 L, i = 2.47, T = (27 + 273)K = 300K, R = 0.0821 L atm K-1mol–1
Molar mass of CaCl2 = 111 g mol -1
π = i CRT
C = π/iRT = 0.75/(2.47×0.0821×300) = 0.0123 M
No of moles of CaCl2 = Concentration x Volume (L) = 0.0123 x 2.5 = 0.0375 mol
Mass of CaCl2 = Mole x Molar mass = 0.0375 x 111 = 3.413 g
Hence, the required amount of CaCl2 is 3.413 g.
53. 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.
Answer: (i) Calculation of Van’t Hoff factor (i):
(ii) Calculation of osmotic pressure (π):
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