Get NCERT Class 12 Chemistry Chapter 1 Solutions with detailed step-by-step answers. These solutions are prepared as per the latest CBSE syllabus.
Learn important concepts like molarity, molality, mole fraction and concentration of solutions in an easy way.
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- ✔ Quick revision notes
Below are the complete solutions for Chapter 1 – Solutions.
Define the term solution. How many types of solutions are formed? Write briefly with example.
A solution is a homogeneous mixture of two or more substances. The component present in larger amount is called the solvent and the component present in smaller amount is called the solute.
Types of solutions:
| Type | Solute | Solvent | Example |
|---|---|---|---|
| Gaseous | Gas | Gas | Air |
| Liquid | Gas | Chloroform vapour in nitrogen | |
| Solid | Gas | Camphor in nitrogen | |
| Liquid | Gas | Liquid | Oxygen in water |
| Liquid | Liquid | Ethanol in water | |
| Solid | Liquid | Glucose in water | |
| Solid | Gas | Solid | Hydrogen in palladium |
| Liquid | Solid | Amalgam | |
| Solid | Solid | Alloy |
Give an example of a solid solution in which the solute is a gas.
A solid solution is formed when a gas dissolves in a solid metal.
Example: Hydrogen gas dissolved in palladium.
Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage
(i) Mole Fraction:
XA = nA nA + nB
(ii) Molality:
m = moles of solute mass of solvent (kg)
(iii) Molarity:
M = moles of solute volume of solution (L)
(iv) Mass Percentage:
Mass % = mass of solute mass of solution × 100
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?
Step 1: Assume mass of solution
Assume 100 g of solution
Mass of HNO3 = 68 g
Molar mass of HNO3 = 63 g mol-1
Step 2: Calculate moles of HNO3
Moles = 68 63 = 1.079 mol
Step 3: Calculate volume of solution
Density = 1.504 g mL-1
Volume = Mass Density = 100 1.504 = 66.49 mL
Convert to litre: 66.49 mL = 0.06649 L
Step 4: Calculate molarity
M = moles volume = 1.079 0.06649
Final Answer: Molarity = 16.2 M
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, what shall be the molarity of the solution?
Step 1: Assume mass of solution
Assume 100 g solution
Mass of glucose = 10 g
Mass of water = 90 g
Step 2: Calculate moles
Molar mass of glucose (C6H12O6) = 180 g mol-1
Moles of glucose = 10 180 = 0.0556 mol
Molar mass of water = 18 g mol-1
Moles of water = 90 18 = 5 mol
Step 3: Calculate molality
m = 0.0556 0.09 = 0.617 m
Step 4: Calculate mole fraction
Total moles = 0.0556 + 5 = 5.0556
Xglucose = 0.0556 5.0556 = 0.011
Xwater = 5 5.0556 = 0.989
Step 5: Calculate molarity
Density = 1.2 g mL-1
Volume = 100 1.2 = 83.33 mL
Convert to litre: 83.33 mL = 0.0833 L
M = 0.0556 0.0833 = 0.667 M
Final Answer:
Molality = 0.617 m
Mole fraction of glucose = 0.011
Mole fraction of water = 0.989
Molarity = 0.667 M
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?
Step 1: Assume moles
Let moles of Na2CO3 = x and NaHCO3 = x
Molar mass of Na2CO3 = 106 g mol-1
Molar mass of NaHCO3 = 84 g mol-1
Total mass = 1 g
106x + 84x = 1
190x = 1
x = 1 190 = 0.00526 mol
Step 2: Write chemical reactions
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
Step 3: Calculate total moles of HCl required
For Na2CO3 → HCl required = 2x
For NaHCO3 → HCl required = x
Total HCl = 2x + x = 3x
= 3 × 0.00526 = 0.01578 mol
Step 4: Calculate volume of HCl
M = moles volume
Volume = 0.01578 0.1 = 0.1578 L
0.1578 L = 157.8 mL
Final Answer: Volume of HCl required = 157.8 mL
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.
Step 1: Calculate mass of solute in first solution
Solute = 25 100 × 300 = 75 g
Step 2: Calculate mass of solute in second solution
Solute = 40 100 × 400 = 160 g
Step 3: Calculate total solute and total mass
Total solute = 75 + 160 = 235 g
Total mass = 300 + 400 = 700 g
Step 4: Calculate mass percentage
Mass % = 235 700 × 100 = 33.57 %
Final Answer: Mass percentage of resulting solution = 33.57 %
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, what is the molarity of the solution?
Step 1: Calculate moles of ethylene glycol
Molar mass of C2H6O2 = 62 g mol-1
Moles = 222.6 62 = 3.59 mol
Step 2: Calculate molality
Mass of solvent (water) = 200 g = 0.2 kg
m = 3.59 0.2 = 17.95 m
Step 3: Calculate molarity
Total mass of solution = 222.6 + 200 = 422.6 g
Volume = 422.6 1.072 = 394.2 mL
Convert to litre: 394.2 mL = 0.394 L
M = 3.59 0.394 = 9.11 M
Final Answer:
Molality = 17.95 m
Molarity = 9.11 M
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3). The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass. (ii) Determine the molality of chloroform in water.
Part (i): Conversion to percent by mass
Step 1: Understand ppm
1 ppm = 1 part of solute in 106 parts of solution
15 ppm = 15 106
Step 2: Convert to percent
Mass % = 15 106 × 100 = 1.5 × 10-3 %
Part (ii): Calculate molality
Step 1: Assume 1 kg solution
Mass of chloroform = 15 mg = 0.015 g
Molar mass of CHCl3 = 119.5 g mol-1
Moles = 0.015 119.5 = 1.26 × 10-4 mol
Step 2: Calculate molality
Mass of solvent ≈ 1 kg
m = 1.26 × 10-4 m
Final Answer:
Percent by mass = 1.5 × 10-3 %
Molality = 1.26 × 10-4 m
What role does molecular interaction play in a solution of alcohol and water?
Step 1: Intermolecular forces in pure liquids
In pure water, strong hydrogen bonding exists between water molecules. Similarly, hydrogen bonding also exists between alcohol molecules.
Step 2: Mixing alcohol and water
When alcohol and water are mixed, some of the hydrogen bonds present in pure liquids break. New interactions formed between alcohol and water molecules are comparatively weaker.
Step 3: Effect on vapour pressure
Due to weaker intermolecular forces, molecules can escape more easily into vapour phase. Hence, vapour pressure increases.
Step 4: Deviation from Raoult’s law
Because of weaker interactions, the solution shows positive deviation from Raoult’s law.
Final Conclusion:
Alcohol–water solution shows positive deviation from Raoult’s law due to weaker intermolecular interactions.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Gas dissolution in liquids is generally an exothermic process (heat is released).
According to Le Chatelier’s Principle, when temperature increases, the equilibrium shifts in the direction that absorbs heat.
Therefore, increasing temperature favors the escape of gas molecules from the solution.
As a result, solubility of gases decreases with increase in temperature.
Final Answer: Gas solubility decreases with increase in temperature because dissolution is exothermic.
State Henry’s law and mention some important applications.
Henry’s Law:
At constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
Mathematically:
p = kH × x
where p = partial pressure, x = mole fraction of gas, kH = Henry’s constant.
Applications:
- ✔ Carbonated drinks: CO₂ is dissolved under high pressure
- ✔ Deep-sea diving: explains “bends” due to dissolved gases
- ✔ High altitude: less oxygen dissolves in blood
Final Answer: Henry’s law relates gas solubility with pressure and has important real-life applications.
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, what shall be the partial pressure of the gas?
Step 1: Use Henry’s law
p ∝ mole fraction ⇒ p₁ / p₂ = x₁ / x₂
Since solvent is same:
p₁ / p₂ = m₁ / m₂
Step 2: Substitute values
p₂ = m₂ m₁ × p₁
= 5.00 × 10-2 6.56 × 10-3 × 1
= 7.62 bar (approx)
Final Answer: Partial pressure = 7.62 bar
What is meant by positive and negative deviations from Raoult’s law? How is the sign of ΔHmix related to positive and negative deviations from Raoult’s law?
Positive Deviation:
When the vapour pressure of a solution is higher than expected from Raoult’s law, it shows positive deviation.
This happens when solute–solvent interactions are weaker than solute–solute and solvent–solvent interactions.
Negative Deviation:
When the vapour pressure is lower than expected, the solution shows negative deviation.
This occurs when solute–solvent interactions are stronger.
Relation with ΔHmix:
- ✔ Positive deviation → ΔHmix is positive (endothermic)
- ✔ Negative deviation → ΔHmix is negative (exothermic)
Final Answer: Positive deviation occurs due to weaker interactions (ΔHmix > 0), while negative deviation occurs due to stronger interactions (ΔHmix < 0).
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?
Step 1: Use relative lowering of vapour pressure
Δp p° = nsolute nsolvent
Step 2: Given
p° = 1.013 bar (normal boiling point of water)
p = 1.004 bar
Δp = 1.013 − 1.004 = 0.009 bar
Δp p° = 0.009 1.013 ≈ 0.00888
Step 3: Assume 100 g solution
Solute = 2 g, Solvent = 98 g
Moles of solvent = 98 18 = 5.44 mol
nsolute 5.44 = 0.00888
nsolute = 0.0483 mol
Molar mass = 2 0.0483 ≈ 41.4 g mol-1
Final Answer: Molar mass ≈ 41.4 g mol-1
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of pure 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?
Step 1: Calculate moles
Molar mass of heptane = 100 g mol-1
Moles of heptane = 26 100 = 0.26 mol
Molar mass of octane = 114 g mol-1
Moles of octane = 35 114 = 0.307 mol
Step 2: Calculate mole fractions
Total moles = 0.26 + 0.307 = 0.567
xheptane = 0.26 0.567 = 0.458
xoctane = 0.307 0.567 = 0.542
Step 3: Apply Raoult’s law
P = x1P°1 + x2P°2
P = (0.458 × 105.2) + (0.542 × 46.8)
P = 48.2 + 25.4 = 73.6 kPa
Final Answer: Vapour pressure = 73.6 kPa
The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.
Step 1: Use relative lowering of vapour pressure
Δp p° = nsolute nsolvent
Step 2: For 1 molal solution
1 mol solute in 1 kg water
Moles of water = 1000 18 = 55.55 mol
Δp p° = 1 55.55 = 0.018
Δp = 0.018 × 12.3 = 0.221 kPa
p = 12.3 − 0.221 = 12.08 kPa
Final Answer: Vapour pressure = 12.08 kPa
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%.
Step 1: Use relative lowering
Δp p° = xsolute
Given: p/p° = 0.80 → Δp/p° = 0.20
xsolute = 0.20
Step 2: Moles of octane
Moles = 114 114 = 1 mol
Step 3: Mole fraction relation
x = n n + 1 = 0.20
n = 0.25 mol
Step 4: Mass
Mass = 0.25 × 40 = 10 g
Final Answer: Mass = 10 g
A solution containing 30 g of non-volatile solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. After adding 18 g of water, the vapour pressure becomes 2.9 kPa. Calculate (i) molar mass of the solute (ii) vapour pressure of pure water.
Step 1: Let molar mass of solute = M
Moles of solute = 30 M
Moles of water = 90 18 = 5 mol
Step 2: Apply Raoult’s law
p° − p p° = n₂ n₁ + n₂
p° − 2.8 p° = 30/M 5 + 30/M
⇒ p° 2.8 = 1 + 6 M …… (i)
Step 3: After adding water
New moles of water = 108 18 = 6 mol
p° − 2.9 p° = 30/M 6 + 30/M
⇒ p° 2.9 = 1 + 5 M …… (ii)
Step 4: Divide (i) by (ii)
2.9 2.8 = 1 + 6/M 1 + 5/M
Solving ⇒ M = 23 g mol-1
Step 5: Calculate vapour pressure (p°)
p° 2.8 = 29 23
p° = 3.53 kPa
Final Answer:
Molar mass = 23 g mol-1
Vapour pressure = 3.53 kPa
A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% glucose in water. (Given: Kf for water = 1.86 K kg mol-1)
Step 1: Understand relation
Depression in freezing point: ΔTf = Kf × m
Step 2: For cane sugar solution
Freezing point of pure water = 273 K
Freezing point of solution = 271 K
ΔTf = 273 − 271 = 2 K
m = ΔTf Kf = 2 1.86 = 1.075 mol kg-1
Step 3: Compare molality
Both solutions are 5% by mass, so molality depends on molar mass.
Cane sugar (C12H22O11) = 342 g mol-1
Glucose (C6H12O6) = 180 g mol-1
Molality ∝ 1 Molar mass
mglucose mcane sugar = 342 180
mglucose = 1.075 × 342 180 = 2.04 mol kg-1
Step 4: Calculate new freezing point
ΔTf = Kf × m = 1.86 × 2.04 = 3.79 K
Freezing point = 273 − 3.79 = 269.2 K
Final Answer:
Freezing point of glucose solution = 269.2 K
Two elements A and B form compounds AB₂ and AB₄. When dissolved in 20 g of benzene, 1 g of AB₂ lowers freezing point by 2.3 K whereas 1 g of AB₄ lowers it by 1.3 K. Kf = 5.1 K kg mol⁻¹. Find atomic masses of A and B.
Concept: ΔTf = Kf × m
For AB₂: Molecular mass = M + 2B
m = 1/M0.02
2.3 = 5.1 × 10.02M₁ ⇒ M₁ = 110
Similarly for AB₄: M₂ = 194
Now solving: A + 2B = 110 A + 4B = 194
Subtract → 2B = 84 → B = 42
A = 110 − 84 = 26
At 300 K, 36 g glucose in 1 L solution has osmotic pressure 4.98 bar. If pressure is 1.52 bar, find concentration.
Formula: π = CRT
π₁ / π₂ = C₁ / C₂
4.98 / 1.52 = C₁ / C₂
C₂ = (1.52 / 4.98) × C₁ = 0.305 C₁
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)
(i) n-hexane and n-octane
Both are non-polar molecules. Hence, the only intermolecular force present is London dispersion force. These forces arise due to temporary dipoles.
(ii) I₂ and CCl₄
Both molecules are non-polar and symmetrical. Therefore, the dominant force is London dispersion force.
(iii) NaClO₄ and water
NaClO₄ is an ionic compound (Na⁺ and ClO₄⁻ ions), and water is a polar molecule. Hence, the interaction between ions and dipole results in Ion–dipole interaction.
(iv) Methanol and acetone
Methanol contains –OH group capable of hydrogen bonding. Acetone has oxygen with lone pairs. Thus, strong interaction is Hydrogen bonding.
(v) CH₃CN and acetone
Both molecules are polar but do not form hydrogen bonds. Therefore, the main interaction is Dipole–dipole interaction.
(i) London dispersion
(ii) London dispersion
(iii) Ion–dipole
(iv) Hydrogen bonding
(v) Dipole–dipole
Based on solute–solvent interactions, arrange the following in order of increasing solubility in n-octane: Cyclohexane, KCl, CH₃OH, CH₃CN
Step 1: Identify nature of solvent
n-octane is a non-polar solvent. According to the rule “like dissolves like”, non-polar solvents dissolve non-polar substances better.
Step 2: Analyze each solute
KCl: Ionic compound (K⁺, Cl⁻). Does not dissolve in non-polar solvent → very low solubility.
CH₃OH (Methanol): Polar molecule with hydrogen bonding. Poor interaction with non-polar solvent → low solubility.
CH₃CN (Acetonitrile): Polar molecule but no hydrogen bonding. Better compatibility than alcohol → moderate solubility.
Cyclohexane: Non-polar molecule. Same nature as n-octane → high solubility.
Step 3: Arrange in increasing order
KCl < CH₃OH < CH₃CN < Cyclohexane
Classify the following compounds as insoluble, partially soluble and highly soluble in water: phenol, toluene, formic acid, ethylene glycol, chloroform, pentanol
Concept: Water is a polar solvent and forms hydrogen bonding. Hence, solubility depends on polarity and ability to form hydrogen bonds.
Highly soluble compounds:
Formic acid (HCOOH) → contains –COOH group (strong hydrogen bonding)
Ethylene glycol (HO–CH₂–CH₂–OH) → contains two –OH groups (very strong hydrogen bonding)
Partially soluble compounds:
Phenol (C₆H₅OH) → has –OH group but large non-polar benzene ring
Pentanol (C₅H₁₁OH) → has –OH group but long carbon chain reduces solubility
Insoluble compounds:
Toluene (C₆H₅CH₃) → non-polar hydrocarbon
Chloroform (CHCl₃) → weak polarity, no hydrogen bonding
Highly soluble → Formic acid, Ethylene glycol
Partially soluble → Phenol, Pentanol
Insoluble → Toluene, Chloroform
If the density of some lake water is 1.25 g mL−1 and it contains 92 g of Na+ ions per kg of water, calculate the molarity of Na+ ions in the lake.
Step 1: Calculate moles of Na⁺
Molar mass of Na = 23 g mol⁻¹
Moles of Na⁺ = 92 23 = 4 mol
Step 2: Calculate volume of solution
Mass of solution = 1000 g
Density = 1.25 g mL⁻¹
Volume = 1000 1.25 = 800 mL = 0.8 L
Step 3: Calculate molarity
M = 4 0.8 = 5 M
If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of CuS in aqueous solution.
Step 1: Dissociation
CuS ⇌ Cu²⁺ + S²⁻
Let solubility = s
Ksp = s × s = s²
s² = 6 × 10⁻¹⁶
s = √(6 × 10⁻¹⁶) = 2.45 × 10⁻⁸ mol L⁻¹
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Formula:
Mass % = mass of solute mass of solution × 100
Mass of solution = 6.5 + 450 = 456.5 g
Mass % = 6.5 456.5 × 100 = 1.42%
Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose is 1.5 mg. Calculate the mass of 1.5 × 10−3 m aqueous solution required for the above dose.
Step 1: Calculate molar mass
M = 311 g mol⁻¹
Step 2: Calculate moles
Moles = 1.5 × 10⁻³ 311 = 4.82 × 10⁻⁶ mol
Step 3: Use molality
kg solvent = 4.82 × 10⁻⁶ 1.5 × 10⁻³ = 3.21 × 10⁻³ kg
Mass = 3.21 g
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Step 1: Formula
M = moles volume (L)
Moles = M × V = 0.15 × 0.25 = 0.0375 mol
Step 2: Molar mass
C₆H₅COOH = 122 g mol⁻¹
Mass = 0.0375 × 122 = 4.575 g
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.
Step 1: Concept
Depression in freezing point depends on the number of particles present in solution (van’t Hoff factor, i).
Step 2: Nature of acids
Acetic acid is a weak acid and ionizes very little.
Trichloroacetic acid is stronger and ionizes more.
Trifluoroacetic acid is the strongest and shows maximum ionization.
Step 3: Conclusion
Greater ionization → more particles → greater depression in freezing point.
Calculate the depression in 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.
Step 1: Calculate molar mass
Molar mass of CH₃CH₂CHClCOOH = 122.5 g mol−1
Step 2: Calculate moles
Moles = 10 / 122.5 = 0.0816 mol
Step 3: Calculate molality
Mass of solvent = 250 g = 0.25 kg
m = 0.0816 / 0.25 = 0.326 m
Step 4: Degree of dissociation
CH₃CH₂CHClCOOH ⇌ CH₃CH₂CHClCOO⁻ + H⁺
For dilute solution, (1 − α) ≈ 1
Ka = Cα²
α = √(Ka / C)
α = √(1.4 × 10−3 / 0.326)
α = 0.065
Step 5: van’t Hoff factor
i = 1 + α = 1.065
Step 6: Depression in freezing point
ΔTf = i × Kf × m
= 1.065 × 1.86 × 0.326
≈ 0.64 K
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.
Step 1: Calculate molar mass
CH₂FCOOH = 12×2 + 1×3 + 19 + 16×2 = 24 + 3 + 19 + 32 = 78 g mol⁻¹
Step 2: Calculate molality
Moles = 19.5 78 = 0.25 mol
Mass of solvent = 500 g = 0.5 kg
Molality (m) = 0.25 0.5 = 0.5 m
Step 3: Use freezing point formula
ΔTf = i × Kf × m
1.0 = i × 1.86 × 0.5
i = 1.0 0.93 = 1.075
Step 4: Degree of dissociation (α)
For acid: i = 1 + α
α = i − 1 = 1.075 − 1 = 0.075
Step 5: Dissociation constant (Ka)
Ka = Cα² 1 − α
For dilute solution: (1 − α) ≈ 1
C = 0.5 mol kg⁻¹
Ka = 0.5 × (0.075)²
= 0.5 × 0.005625 = 2.81 × 10⁻³
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.
Step 1: Calculate moles of solute (glucose)
Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹
Moles of glucose = 25 180 = 0.139 mol
Step 2: Calculate moles of solvent (water)
Moles of water = 450 18 = 25 mol
Step 3: Apply Raoult’s law
P = Xsolvent × P°
Xwater = 25 25 + 0.139 = 0.9945
Step 4: Calculate vapour pressure
P = 0.9945 × 17.535
P ≈ 17.44 mm Hg
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.
Step 1: Use Henry’s Law
p = KH × x
x = p KH
Step 2: Substitute values
x = 760 4.27 × 10⁵
x = 1.78 × 10⁻³
Step 3: Interpretation
This value represents mole fraction (solubility) of methane in benzene.
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.
Step 1: Calculate moles of A and B
Moles of A = 100 140 = 0.714 mol
Moles of B = 1000 180 = 5.56 mol
Step 2: Calculate mole fractions
XA = 0.714 0.714 + 5.56 = 0.114
XB = 1 − 0.114 = 0.886
Step 3: Vapour pressure of B in solution
PB = XB × P°B
PB = 0.886 × 500 = 443 torr
Step 4: Vapour pressure of A in solution
PA = Total − PB
PA = 475 − 443 = 32 torr
Step 5: Vapour pressure of pure A
PA = XA × P°A
P°A = 32 0.114 = 281 torr
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 mixture is:
Step 1: Given Data
| 100 × xacetone | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
|---|---|---|---|---|---|---|---|---|
| pacetone | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| pchloroform | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
| ptotal | 632.8 | 603.0 | 579.5 | 562.1 | 580.4 | 599.5 | 615.3 | 641.8 |
Step 2: Ideal solution (Raoult’s Law)
pacetone = xacetone × 741.8
pchloroform = (1 − xacetone) × 632.8
ptotal = pacetone + pchloroform
Step 3: Graph observation
Ideal solution → straight lines Experimental curve → bends downward
Step 4: Conclusion
Since actual vapour pressure is lower than ideal, the solution shows negative deviation.
Reason: Strong hydrogen bonding between acetone and chloroform.
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.
Step 1: Calculate moles
Molar mass of benzene (C₆H₆) = 78 g mol⁻¹
Moles of benzene = 80 78 = 1.026 mol
Molar mass of toluene (C₇H₈) = 92 g mol⁻¹
Moles of toluene = 100 92 = 1.087 mol
Step 2: Mole fractions in liquid phase
Xbenzene = 1.026 1.026 + 1.087 = 0.486
Xtoluene = 1 − 0.486 = 0.514
Step 3: Apply Raoult’s Law
pbenzene = 0.486 × 50.71 = 24.64 mm Hg
ptoluene = 0.514 × 32.06 = 16.48 mm Hg
Step 4: Total vapour pressure
ptotal = 24.64 + 16.48 = 41.12 mm Hg
Step 5: Mole fraction in vapour phase
ybenzene = pbenzene ptotal
= 24.64 41.12 = 0.599
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% and 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 are 3.30 × 10⁷ mm Hg and 6.51 × 10⁷ mm Hg respectively, calculate the composition of these gases in water.
Step 1: Convert pressure into mm Hg
Total pressure = 10 atm = 10 × 760 = 7600 mm Hg
Step 2: Partial pressures of gases
pO₂ = 0.20 × 7600 = 1520 mm Hg
pN₂ = 0.79 × 7600 = 6004 mm Hg
Step 3: Apply Henry’s Law
x = p KH
For Oxygen:
xO₂ = 1520 3.30 × 10⁷ = 4.61 × 10⁻⁵
For Nitrogen:
xN₂ = 6004 6.51 × 10⁷ = 9.22 × 10⁻⁵
Step 4: Conclusion
Nitrogen is more soluble than oxygen in water under given conditions.
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.
Step 1: Use osmotic pressure formula
π = i × C × R × T
C = π iRT
Step 2: Substitute values
π = 0.75 atm, i = 2.47, R = 0.0821 L·atm·mol⁻¹·K⁻¹, T = 300 K
C = 0.75 2.47 × 0.0821 × 300
C = 0.0123 mol L⁻¹
Step 3: Calculate moles
Moles = C × V = 0.0123 × 2.5 = 0.0308 mol
Step 4: Calculate mass
Molar mass of CaCl₂ = 111 g mol⁻¹
Mass = 0.0308 × 111 = 3.42 g
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.
Step 1: Convert mass into moles
Mass = 25 mg = 0.025 g
Molar mass of K₂SO₄ = 174 g mol⁻¹
Moles = 0.025 174 = 1.44 × 10⁻⁴ mol
Step 2: Calculate molarity
Volume = 2 L
C = 1.44 × 10⁻⁴ 2 = 7.2 × 10⁻⁵ mol L⁻¹
Step 3: Van’t Hoff factor
K₂SO₄ → 2K⁺ + SO₄²⁻ i = 3
Step 4: Use osmotic pressure formula
π = i × C × R × T
π = 3 × 7.2 × 10⁻⁵ × 0.0821 × 298
π ≈ 5.27 × 10⁻³ atm
