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NCERT Class 12 Chemistry Chapter 2 Solutions
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Solutions
Chapter: 2
| Part – I |
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 where a solute is uniformly dissolved in a solvent. Solutions can be categorised into different types based on the states of the solute and solvent:
(i) Gas in gas: When one gas is combined with another, it forms a gas-in-gas solution.
Example: Air is a mixture of nitrogen and oxygen.
(ii) Solid in Gas: When tiny solid particles are suspended in a gas, it creates a solid-in-gas solution.
Example: Moisture (water in air).
(iii) Liquid in Gas: When liquid is mixed with a large amount of gas, it is called liquid in gas solution.
Example: Moisture (water droplets in the air)
(iv) Liquid in liquid: Where a liquid is miscible with another liquid, it forms a solution of liquid in liquid.
Examples: Ethanol mixed with water, methanol blended with water.
2. Give an example of a solid solution in which the solute is a gas.
Ans: When a solid solution is created between two substances, with one having very large particles and the other having very small particles, an interstitial solid solution is formed. For instance, a solution of hydrogen in palladium is a solid solution where the solute is a gas.
An example of a solid solution in which the solute is a gas is hydrogen dissolved in palladium.
3. Define the following terms:
(i) Mole fraction.
Ans: Mole Fraction: The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.
(ii) Molality.
Ans: Molality: It is the number of moles of the solute per kilogram of the solvent. It is the ratio of the number of moles of solute to the mass of solvent in kg.
(iii) Molarity.
Ans: Molarity (M): The concentration expressed as the moles of solute per litre of solution is the molarity. It is denoted by M.
(iv) Mass percentage.
Ans: Mass Percentage: The mass percentage of a component in a solution is defined as the mass of the solute in grams present in 100 grams of the solution.
4. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1
Ans: 68 % by mass implies that 68 g of HNO3 are present in 100g of solution.
Volume of solution = Mass of solution / Density of solution
100 g/ (1.504 g mL-1)
= 66.5 cm3
= 0.665 L.
The molarity of a solution represents the number of moles of nitric acid present in 1 litre of the solution.
MB of NHO3 = 1 + 14 + 48 = 63.
Hence, the molarity should be 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: Here,
Glucose solution: 10% w/w (weight/weight), meaning 10 g of glucose is dissolved in 90 g of water (total mass = 100 g).
Density of solution: 1.2 g/mL.
Molar mass of glucose (C₆H₁₂O₆): 180 g/mol.
Molar mass of water (H₂O): 18 g/mol.
Mass of water = 100 -10
= 90 g
= 0.90 kg
10g glucose = 10 / 180 mol
= 0.0555 mol
Number of moles in 90g H2O
90 / 18
= 5 moles
Molality of solution = 0.0555 mol / 0.090
= 0.617m.
Mole fraction of glucose
= 0.0555 / 5 + 0.0555 = 0.01
Moles fraction of H2O = 1 – 0.01 = 0.99
Volume of 100g solution
= 100 / 1.2 ml
= 83.33 ml
= 0.08333 L
∴ Molarity = 0.0555 / 0.08333
= 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: Total mass = 1 g
Let the mass of Na₂CO₃ = x g
Mass of NaHCO₃ = 1−x g
The number of moles of sodium carbonate is given as x/106,
The number of moles of sodium bicarbonate is (1−x)/84.
It is an equimolar mixture,
= Mass / Molar mass
= x/106
= (1-x) / 84 mol
= x/106 mol = (1-x) / 84 mol
84x = 106-106x
106x + 84x = 106
190x = 106
X = 106 / 190 = 0.558g
Mass of NaCO3 = 0.558g
Mass of NaHCO3 = (1-0.558)
= 0.442g
Calculating total mass Of HCI required
Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol
Total moles of HCl = 0.01578 mol
Mass of HCl = 0.01578 mol × 36.5 g/mol = 0.576 g
Calculate volume of 0.1 M HCl:
V = 0.01578 mol0.1 M = 0.1578 L
=157.8 mL
So, 157.8 mL of 0.1 M HCl is required.
7. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Ans: Mass of solute in 300 g of solution
= 300 – 75 = 225g
Total Mass of the Solution:
= 300g + 400g = 700g
Mass of the Solute in Each Solution
For the 25% solution
Mass of solute
= 0.25 × 300g = 75g
For the 40% solution:
Mass of solute = 0.40 × 400 g = 160 g
Mass Percentage of the Resulting Solution
Mass percentage = (235g/ 700g ) × 100 = 33.6%
Total mass of solvent
= 225g + 240g = 465g
Percentage composition of solvent in solution after mixing
= 465 / 700
= 66.43%
Therefore, composition of solute in solution after mixing
= 235 / × 100
= 33.57 %
8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2 ) and 200 g of water. Calculate the morality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?
Ans: Moles of ethylene glycol (C₂H₆O₂):
Moles = 222.6 g /62.07 g
= 62.07 g
Mass of water (solvent):
Mass=200 g=0.200 kg
Molality (m)
Molality = 3.58 mol / 0.200 kg
= 17.9 mol / Kg.
= 17.95 m
Total mass of solution
= 222.6 + 200 = 422.6 g
Density of solution
= 1.672 g/ ml
Volume of solution (L):
Mass / Density = 422.6 / 1.072
= 394.2 ml
= 0.3942 L
Molarity (M):
=3.58 mol / 0.394L
= 9.10mol/L;
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: 15 ppm of CHCl in water means that the is 15 g of CHCl3, in 106 g of water (1 million = 106) Percentage of CHCI
Convert to percent by mass:
Percent by mass = 15g / 106 × 100
= 0.0015 %
= 105 × 10-4 %
(ii) determine the molality of chloroform in the water sample.
Mass of CHCl₃ = 15 mg = 0.015 g
Molar mass of CHCl₃ = 119.5 g/mol
Mass of water = 1 kg = 1000 g
= moles of CHCL3 / mass of water in kg
M = 15/ 119.5 / 103
Molality (m)
= 1.41 × 10-4.
10. What role does molecular interaction play in a solution of alcohol and water?
Ans: Between the molecules of alcohol and water, hydrogen bonding is a key interaction. When alcohol and water are mixed, new hydrogen bonds form between the alcohol and water molecules, but these bonds are weaker than the hydrogen bonds within each pure substance. As a result, the overall strength of intermolecular attractions decreases. This will lead to an increase in vapour pressure of the solution and a decrease in its boiling point.
11. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Ans: The dissolution of a gas in a liquid is an exothermic process, meaning that heat is released during the dissolution. (Gas + Solvent ⇌ Solution + Heat)
As the temperature is increased, equilibrium shifts backward, according to Le Chateliar’s Principle
12. State Henry’s law and mention some important applications.
Ans: Henry’s Law states that “the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.”
It is expressed as
P = KHX⋅
Where,
KH = is Henry’s Law constant.
p.= is the partial pressure of the gas in the vapour phase.
X= Molar fraction of the gas.
Application of Henry’s law:
(i) In packing of soda cans: Soda water bottles are always packed under higher pressure to increase the solubility of CO2 gas.
(ii) To avoid the toxic effects of high concentration of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 50.2% nitrogen and 32.1% oxygen)
(iii) At high altitudes, low blood oxygen levels can cause climbers to feel weak and have difficulty thinking clearly, symptoms of a condition known as hypoxia.
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: According to Henry’s law, the solubility of gas in a liquid is directly proportional to the pressure of the gas.
M = KH × p
In the first case,
= 6.50 × 10-2g bar
Let the partial pressure in the second case be ppp. In this scenario, the amount of ethane is 5.00 × 10⁻² g.
= (6.50 × 12 g bar-1) × p
(KH remains the same)
Or
= 0.762 bar.
14. What is meant by positive and negative deviations from Raoult’s law and how is the sign of Dmix related to positive and negative deviations from Raoult’s law?
Ans: Positive Deviation from Raoult’s law. In those non-ideal solutions, when partial pressure of component ‘A’ in the mixture of ‘A’ and ‘B’ is found to be more than that calculated from Raoult’s law./ Similarly, the partial vapour pressure of component ‘B’ can be higher than calculated from Raoult’s law. This type of deviation from ideal behaviour is called positive deviation from Raoult’s law, e.g., water and ethanol, chloroform and water, ethanol and CCI, methanol and chloroform, benzene and methanol, acetic acid and toluene, acetone and ethanol, methanol and H,O.
For positive deviation AH > 0. (+ve) mixing
When the partial vapour pressure of component ‘A’ is lower than predicted by Raoult’s law after adding component ‘B’, the solution exhibits a negative deviation from ideal behaviour. This means the actual vapour pressure is lower than that of an ideal solution with the same composition, leading to a higher boiling point compared to the individual components. Examples include mixtures like chloroform and acetone, chloroform and methyl acetate, water and HCl, water and HNO₃, acetic acid and pyridine, and chloroform and benzene.
For negative deviation Delta H mixirig <0
15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Ans: Applying Roault’s law for dilute solution (being 2%)
PoA = Vapour pressure of pure solvent 1.013 bar
Ps = vapour pressure of the solution 1.004 bar
WB = mass of solute 2g
= mass of solvent (water) 98g
WA = molar mass of water 18g mol
= 41.0 g 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 26.0 g of heptane and 35 g of octane?
Ans: Number of moles of octane:
(nA) = Mass / Molar mass
For 35 g of octane:
= 35g / 114g mol-1
= 0.307 mol
Molar mass of octane, C8H18
Molar mass = (12 × 8) + (1 × 18)
= 114g mol-1
Number of moles of heptane:
(nB) = 26 G / 100 g mol-1
= 0.26 mol
Molar fraction of octane
Mole fraction of heptane,
(x A) = nB / (nA + nB)
Vapour pressure of pure heptane (Pheptane∘):
(pB) = 105.2 kpa
Vapour pressure of pure octane
(PA) = 46.8 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: The vapour pressure of water is 12.3 kPa at 300 K. We need to calculate the vapour pressure of a 1 molal solution of a non-volatile solute in it.
Or 12.3 – PA / 12.3 = 1 / 56.55
12.3 – PA = 12.3 / 56.55
PA = 12.3 – (12.3 / 56.55)
= 12.3 – 0.0175
= 12.0825 kPa
= 12.08 kPa.
18. Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Ans: We have the relative lowering of vapour pressure.
Or 100-80 / 100 =
19. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
(i) molar mass of the solute.
(ii) vapour pressure of water at 298 K.
Ans: let the molecular mass of the solute be x g per mole
Component of A (water)
WA = 90 g
mA = 18g
Component of B (non-Volatile solute)
WB = 30g
Molar mass = x
nA = 90 / 18 = 5 moles
nB = 30 / x moles
Now PA = P A ×
20. A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.
Ans: Given:
Freezing point of 5% cane sugar solution (C12H22O11) = 271 K
Freezing point of pure water = 273.15 K
Molar mass of cane sugar (C12H22O11) = 342 g/mol
Molar mass of glucose (C6H12O6) = 180 g/mol
Molarity of sugar solution
= (5/342) × (1000/100)
= 0.146
Calculation of depression in freezing point for cane sugar:
273.15-27=2.15
The molality of the cane sugar:
Mass of cane sugar in 100 g solution = 5 g
Mass of water in 100 g solution = 95 g
Molality (mmm) of cane sugar solution:
Molality = moles of solute/kg of solvent
= 5g/(342g/mol) × (0.095kg)
Calculate the molality of a 5% glucose solution:
Mass of glucose in 100 g solution = 5 g
Mass of water in 100 g solution = 95 g
Molality= 5g/(180g/mol × 0.095kg
Freezing point of glucose solution
= 273.15 – 4.09
= 269.06k.
21. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6 ), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B.
Ans: Let ‘a’ be the atomic mass of element A and ‘b’ be the atomic mass of element B.
For compound AB2
1 g of compound AB2 in 20 g of benzene means that 1000 g of benzene contains 50 g of AB2 .. Molality of AB2 in benzene
Given:
ΔTf = 2.3K
WB =1.0g,
WA = 20.0g,
Kf = 5.1K kg/mol
MAB2 = (Kf × WB × 1000)/(wa × ΔTf)
MAB2 = (5.1 × 1.0 × 1000)/ (20.0 × 2.3)
= 110.87g/mol
For compound AB4:
Given:
ΔTf = 1.3K,
WB = 1.0g,
WA = 20.0g
MAB4 = (5.1 × 1.0 × 1000)/(20.0 × 1.3)
=196.15g/mol.
Substituting (i) and (ii), we get
(a + 4b) − (a + 2b) = 196.15 − 110.87
= 2b = 85.28
B = 85.28/2
= 42.64g/mol
Substituting the value of b in (i), we get
a + 2 × 42.64 = 110.87
a + 85.28 = 110.87
a = 110.87 − 85.28
= 25.59g/mol
Thus, atomic mass of element is a
= 25.59 and that of element B is b
= 42.64.
22. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Ans: Given:
mass of glucose = 36 g
Molar mass of glucose = 180 g/mol
Volume of solution = 1 L
C1 = molar mass/mass
= l36g/180g/mol
= 0.2mol/L.
𝝅1 = 4.98 bar
𝝅2 = 1.52 bar
Dividing Equation (ii) by Equation (i):
𝝅2/𝝅1 = C2/C1
⇒ 1.52/4.98
= C2/0.2
C2 = (1.52/4.98) × 0.2
= 0.061M
Thus, the concentration of solution,
C = 0.061 moles/litre.
23. Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane.
Ans: Van der Waals Interactions.
(ii) I 2 and CCl4
Ans: Ion-Dipole Interactions.
(iii) NaClO4 and water.
Ans: Ion dipole interactions.
(iv) methanol and acetone.
Ans: Hydrogen Bonding.
(v) acetonitrile (CH3CN) and acetone (C3H6O).
Ans: 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) Cyclohexane and n-octane both are non- polar. So, they mix completely in all proportions.
(ii) KCl is an ionic compound, but n-octane is non- polar. So, KCl does not dissolve in n-octane.
Although both C3HOH (propanol) and C3H7CN (propanenitrile) are polar, C3H7CNis less polar compared to C3H7OH. In a non-polar solvent like n-octane or hexane, C3H7CN will dissolve more readily than C3H7OH.
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: Insoluble: (ii) Toluene, (v) Chloroform.
Partially Soluble: (i) Phenol, (vi) Pentanol.
Highly Soluble: (iii) Formic Acid, (iv) Ethylene Glycol.
26. If the density of some lake water is 1.25g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molarity of Na+ ions in the lake.
Ans: Mass of sodium ions = 92 g
Molar mass of sodium ions = 23 g/mol
Mass of water = 1 kg
Number of gm moles of solute = 92/23 = 4
Molality = 4/1 = 4m.
27. If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
Ans: CuS (s) ↔ Cu2++S2−
Suppose solubility of CuS is x mol-1
This would give x mol-1 of Cu2+ ions and x mol L-1 of S2- ions on dissociation.
[Cu2+] = xmol-1
[S2+] = xmol-1
Ksp = [Cu2+] [S2+]
= (x) (x) = x2
Ksp = 6 × 10−16
6 × 10−16 = x2
X = √6 × 10−16
X = √10−8
2.45 × 10−8 M.
Maximum molarity = 2.45 × 10−8 M
28. Calculate the mass percentage of aspirin (C9H8O4 ) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Ans: Mass of aspirin (solute) = 6.5 g
Mass of acetonitrile (solvent) = 450 g
Mass percentage = (Mass of solute/ Mass of solute+Mass of solvent) ×100
Total mass of the solution = 6.5g + 450g = 456.5g
Mass% = (6.5/456.5) × 100
= 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: Nalorphene: 1.5 mg = 1.5 × 10−3
MB = 19 × 12 + 21 × 1 + 14 + 3 × 16 = 311 g/mol
m = (WB/MB × wA) ×1000
here:
wB = mass of the solute (nalorphene)
MB = molar mass of the solute
wA = mass of the solvent in kg
1.5 × 10−3 = (1.5 × 10−3/311 × wA) × 1000
wA = (1.5 × 10−3/311 × 1.5 × 10−3) × 1000
wA =1000/311 = 3.2g
30. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Ans: Number of moles = Molarity × Volume in litres
= 0.15 × (250/1000)
= 0.0375 moles
Mol. mass of benzoic acid,C6H5COOH
= 6 × 12 + 5 × 1 + 12 + 16 16 + 1
= 72 + 5 +12 +16 +16 1
= 122 g mol-1
Mass = Molar mass×Number of moles
= 122 × 0.0375
= 4.575 g
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: When strongly electron withdrawing groups are present on alpha C atom of acetic acid, the acid strength and the degree of dissociation increases. The van’t Hoff factor iii increases, causing greater freezing point depression. Trifluoroacetic acid is the most acidic due to fluorine’s strong electron-withdrawing properties.
Hence, for some amount of acetic, thichloro acetic avid and trifluoroacetic acid increases in the given order.
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.
Ans: Calculate the molar mass of CH3CH2CHClCOOH
Molar mass = 12 + 3 + 12 + 2 + 12 + 1 × 35.5 + 12 + 16 + 16 + 1
= 122.5g mol -1
ΔTf = Kf × w2 × 1000
33. 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Ans: Mass of CH₂FCOOH = 19.5 g
Molar mass of CH₂FCOOH = 78 g/mol
Mass of water = 500 g
Freezing point depression (ΔTf) = 1.0°C
Kf for water = 1.86 K kg mol⁻¹
ΔTf = (W × 1000 × KF)(Wm)
1.0 = (19.5 × 1000 × 1.86)/(500.78)
= Van’t Hoff factor
i = (Observed Colligative Property)/(Calculated Colligative Property)
1/0.93 = 1.0753
Let α be the degree of dissociation.
i = (No. of moles present at equilibrium)/(No.of moles taken)
1.073 = 1 + α /1
1 + α = 1.0753
Α = 1.0753 − 1 = 0.0753
C = 39g/78g/mol
= 0.5 mol/kg
= Ka = (0.50 × 0753)2)/ (1−0.0753)
=0.5 × 0.753 × 0.753/(0.925)
= 0.0028125/0.925
= 0.00304 = 3.0 × 10-3.
34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Ans: Vapour pressure of pure water (Pwater0A) at 293 K = 17.535 mm Hg
Mass of glucose = 25 g
Mass of water = 450 g
Molar mass of glucose (C6H12O6C) = 180 g/mol
Molar mass of water = 18 g/mol
Ps = ?
P8 = 17.535 – 0.097
= 17.438 mm 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: According to Henry’s law
p = RHx
kH = 4.27 × 105 mm Hg,
p = 760 mm
760 = 4.27 × 105 × x
Here, C represents the molality of methane in benzene, p denotes the pressure of methane, and k stands for Henry’s law constant.
36. 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.
Ans: Mass of liquid A = 100 g
Molar mass of liquid A = 140 g/mol
Mass of liquid B = 1000 g
Molar mass of liquid B = 180 g/mol
Vapour pressure of pure liquid B (P0B) = 500 Torr
Total vapour pressure of the solution = 475 Torr
WB = 100g, nA
= 100 / 40 = 2.5
WA = 1000g, nB
= 180 / 1000 = 0.18
Or Po = 475 / 14.9 = 31.88 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 an ideal solution over the entire range of composition, plot p total , p chloroform , and p acetone as a function of x acetone . The experimental data observed for different compositions of mixture is:
| 100 x xacetone | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| pacetone /mm Hg | 0 | 54.9 1 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| P chloroform /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 x xacetone | 0.0 | 0.1181 | 0.234 | 0.360. | 0.5088 | 0.582 | 0.645 | 0.721 |
| pacetone /mm Hg | 0 | 54.9 1 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| P chloroform /mm Hg | 632.8 | 548.1 | 469.4 | 359.7 | 259.7 | 193.6 | 162.2 | 120.7 |
| P Total | 632.8 | 603.0 | 579.5 | 562.1 | 580.4 | 599.5 | 615.3 | 641.8 |
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 80 g of benzene is mixed with 100 g of toluene.
Ans: Given:
Mass of benzene (m benzene) = 80 g
Molar mass of benzene (M benzene) = 78 g/mol
Mass of toluene (m toluene) = 100 g
Molar mass of toluene (M toluene) = 92 g/mol
Vapour pressure of pure benzene (P benzene) = 50.71 mm Hg
Vapour pressure of pure toluene (Ptoluene = 32.06 mm Hg
The number of moles
Moles of benzene:
nB = l80g / 78gmo
= 1.026mol
Moles of toluene:
nT = 100g /( 92g / mol )
= 1.087mol
Mole fraction of benzene:
Mola fraction of Naphthalene:
Applying Raoult’s law
Partial pressure of benzene:
B = POB × XB = 50.71 × 0.486
= 24.65mm Hg
Partial pressure of toluene:
PT = PoT × XT = 32.06 × 0.514
= 16.48mm Hg
Total vapour pressure = 24.65 + 16.48 = 41.13 mm Hg
Mole fraction of benzene in vapour phase:
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 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.
Ans: Here:
Oxygen proportion by volume = 20% = 0.20
Nitrogen proportion by volume = 79% = 0.79
Henry’s law constant for oxygen (KH,O2) = 3.30 × 107 mm Hg
Henry’s law constant for nitrogen (KH,N2K) = 6.51 × 107 mm Hg
Calculation of partial pressure of oxygen and nitrogen
Partial pressure of O2
(Po2) = (10atm) × 20/100 = 2atm
= 2 × 760 mm
Partial pressure of N2
(PN2) = ( 10 atm) × 79 / 100
= 79 × 760 mm.
Calculate the partial pressures of oxygen and nitrogen:
xO2 = PO2/ KH,O2
= 1520 / (3.30 × 107)
= 4.61 × 10−5
xN2= PN2 / (KH,N2))
= 6004 / (6.51× 107)
= 9.22 × 10−5
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: According to Van’t Hoff equation
π = osmotic pressure = 0.75 atm
i = Van’t Hoff factor = 2.47 for CaCl2
C = concentration in mol/L
T = temperature in Kelvin = 27∘C + 273 = 300 K
R = gas constant = 0.0821 L atm K⁻¹ mol⁻¹
Volume = 2.5 L
nB = πV / iRT
= (0.0308 mol) × (111 gmol-1)
= 3.42g.
Amount of CaCl2 dissolved = nB
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: Given:
Mass of K₂SO₄ = 25 mg = 0.025 g
Volume of solution = 2 L
Temperature, T = 25°C = 298 K
Molar mass of K₂SO₄ = 174 g/mol
Complete dissociation of K₂SO₄ gives 3 ions (2K⁺ and SO₄²⁻)
Amount of K2SO4 dissolved
= 5mg = 0.025g
Volume of solution
= 2L
T = 25°C
= 25 + 273 K = 298 K
Molar mass of K2SO4
(2 × 39 ) + (32) + (4 × 16)
= 174g mol-1
Since K,SO, dissociates completely as
Total ions produced after dissociation (per moles) = 3
So i = 3
p = iCRT
= i × n / V (RT)
= 5.27 × 10-3 atm.
