SaitechAI – Top 50 Numericals
Chapter: Solutions (Class 12)

Practice set: 50 board-style numericals from NCERT “Solutions”. Click Answer to toggle the worked solution.

A. Basic Concentration & Dilution

Q1. Calculate the mass percentage of glucose in a solution made by dissolving 18 g of glucose in 82 g of water.

Mass of solution = 18 g + 82 g = 100 g
Mass % of glucose = (18 / 100) × 100 = 18% w/w.

Q2. Calculate the mass percentage of NaOH in a solution containing 5 g of NaOH in 45 g of water.

Mass of solution = 5 + 45 = 50 g
Mass % NaOH = (5 / 50) × 100 = 10% w/w.

Q3. Find the molarity of a solution obtained by dissolving 4.6 g of NaCl (M = 58.5 g mol⁻¹) in enough water to make 500 mL of solution.

Moles NaCl = 4.6 / 58.5 ≈ 0.0786 mol
Volume = 0.500 L
Molarity, M = 0.0786 / 0.500 ≈ 0.16 mol L⁻¹.

Q4. Calculate the molarity of a solution containing 49 g of H₂SO₄ (M = 98 g mol⁻¹) per 500 mL of solution.

Moles H₂SO₄ = 49 / 98 = 0.5 mol
Volume = 0.500 L → M = 0.5 / 0.5 = 1.0 mol L⁻¹.

Q5. Calculate the molality of a solution containing 5 g of urea (M = 60 g mol⁻¹) in 100 g of water.

Moles urea = 5 / 60 ≈ 0.0833 mol
Mass of solvent = 0.100 kg
Molality, m = 0.0833 / 0.100 ≈ 0.83 mol kg⁻¹.

Q6. Find the molality of a solution obtained by dissolving 36.5 g of HCl (M = 36.5 g mol⁻¹) in 500 g of water.

Moles HCl = 36.5 / 36.5 = 1 mol
Mass of solvent = 0.500 kg
m = 1 / 0.5 = 2.0 mol kg⁻¹.

Q7. Calculate the mole fraction of ethanol in a solution containing 23 g of ethanol (M = 46 g mol⁻¹) and 36 g of water (M = 18 g mol⁻¹).

n(ethanol) = 23 / 46 = 0.5 mol
n(water) = 36 / 18 = 2 mol
Total moles = 2.5 mol
Mole fraction of ethanol, Xₑ = 0.5 / 2.5 = 0.20.

Q8. Calculate the mole fraction of KCl in a solution containing 10 g of KCl (M = 74.5 g mol⁻¹) in 90 g of water (M = 18 g mol⁻¹).

n(KCl) = 10 / 74.5 ≈ 0.134 mol
n(H₂O) = 90 / 18 = 5.0 mol
Total = 5.134 mol
X(KCl) = 0.134 / 5.134 ≈ 0.026.

Q9. A solution contains 0.5 g of solute in 1.0 kg of solution. Express the concentration in ppm (parts per million).

ppm = (mass of solute / mass of solution) × 10⁶
= (0.5 / 1000) × 10⁶ = 500 × 10³ / 10³ = 500 ppm.

Q10. What will be the normality of 0.20 M H₂SO₄ for acid–base reactions?

H₂SO₄ is dibasic → n-factor = 2
N = M × n-factor = 0.20 × 2 = 0.40 N.

Q11. What volume of 0.50 M NaOH solution contains 0.10 mol of NaOH?

M = n / V → V = n / M = 0.10 / 0.50 = 0.20 L = 200 mL.

Q12. 500 mL of a 2.0 M solution is diluted to 1.0 L. Find the new molarity.

M₁V₁ = M₂V₂
2.0 × 0.500 = M₂ × 1.0 → M₂ = 1.0 mol L⁻¹
New molarity = 1.0 M.

B. Henry’s Law & Gas Solubility

Q13. For a gas in a liquid, the Henry’s law constant KH = 1.5 × 10⁵ atm. Calculate the mole fraction of the gas in the solution when the partial pressure of the gas is 3 atm.

Henry’s law: p = KH x → x = p / KH
x = 3 / (1.5 × 10⁵) = 2.0 × 10⁻⁵
Mole fraction = 2.0 × 10⁻⁵.

Q14. KH for a gas is 2.0 × 10⁵ atm at a certain temperature. What is the mole fraction of the gas in water if its partial pressure is 5.0 atm?

x = p / KH = 5.0 / (2.0 × 10⁵) = 2.5 × 10⁻⁵
Mole fraction = 2.5 × 10⁻⁵.

Q15. The mole fraction of a gas in water is 5.0 × 10⁻⁴ and KH = 2.5 × 10⁵ atm. Calculate the partial pressure of the gas above the solution.

p = KH x = 2.5 × 10⁵ × 5.0 × 10⁻⁴ = 125 atm
Partial pressure = 1.25 × 10² atm.

C. Raoult’s Law & Vapour Pressure

Q16. Vapour pressure of pure water at 25 °C is 23.8 mm Hg. Relative lowering of vapour pressure on adding a non-volatile solute is 0.020. Calculate the vapour pressure of the solution.

p = p⁰ (1 − Δp/p⁰) = 23.8 × (1 − 0.020) = 23.8 × 0.98 ≈ 23.3 mm Hg
Vapour pressure ≈ 23.3 mm Hg.

Q17. For a solution of a non-volatile solute, the relative lowering of vapour pressure is 0.030. What is the mole fraction of the solute?

For non-volatile solute: Δp/p⁰ = Xsolute
So Xsolute = 0.030
Mole fraction = 0.030.

Q18. An ideal binary solution has PA⁰ = 200 mm Hg and PB⁰ = 100 mm Hg. If XA (liquid) = 0.60, calculate the total vapour pressure.

XB = 1 − 0.60 = 0.40
p = XAPA⁰ + XBPB⁰ = 0.60×200 + 0.40×100 = 120 + 40 = 160 mm Hg
Total vapour pressure = 160 mm Hg.

Q19. Using data of Q18, calculate the partial pressure of component B.

pB = XBPB⁰ = 0.40 × 100 = 40 mm Hg.

Q20. Using the same solution as Q18, calculate the mole fraction of A in the vapour phase.

pA = XAPA⁰ = 0.60×200 = 120 mm Hg
Total p = 160 mm Hg (from Q18)
yA = pA / p = 120 / 160 = 0.75
Mole fraction of A in vapour = 0.75.

Q21. 1 mol of a non-volatile solute is dissolved in 9 mol of a solvent. Calculate the relative lowering of vapour pressure.

Mole fraction of solute = 1 / (1 + 9) = 0.10
Δp/p⁰ = Xsolute = 0.10.

Q22. 0.5 mol of a non-volatile solute is dissolved in 9.5 mol of solvent. Find the relative lowering of vapour pressure.

Total moles = 0.5 + 9.5 = 10
Xsolute = 0.5 / 10 = 0.05
So Δp/p⁰ = 0.05.

D. Elevation in Boiling Point & Depression in Freezing Point

Q23. Calculate the elevation in boiling point of water for a 2.0 molal aqueous solution. (Kb for water = 0.52 K kg mol⁻¹)

ΔTb = Kb m = 0.52 × 2.0 = 1.04 K.

Q24. A 0.20 molal aqueous solution is prepared. Find the depression in freezing point. (Kf for water = 1.86 K kg mol⁻¹)

ΔTf = Kf m = 1.86 × 0.20 = 0.372 K.

Q25. Boiling point of pure water is 373 K. What will be the boiling point of a 0.50 molal solution (Kb = 0.52 K kg mol⁻¹)?

ΔTb = 0.52 × 0.50 = 0.26 K
New boiling point = 373 + 0.26 = 373.26 K.

Q26. 1.0 g of a non-volatile solute is dissolved in 100 g of water and the boiling point is raised by 0.512 K. Calculate the molar mass of the solute. (Kb for water = 0.512 K kg mol⁻¹)

m = ΔTb / Kb = 0.512 / 0.512 = 1.0 mol kg⁻¹
For 0.100 kg solvent: moles solute = 1.0 × 0.100 = 0.10 mol
Molar mass = mass / moles = 1.0 / 0.10 = 10 g mol⁻¹.

Q27. 2.0 g of a non-volatile solute is dissolved in 50 g of benzene. The boiling point is elevated by 0.84 K. Calculate the molar mass of the solute. (Kb for benzene = 2.8 K kg mol⁻¹)

m = ΔTb / Kb = 0.84 / 2.8 = 0.30 mol kg⁻¹
Solvent mass = 0.050 kg → moles solute = 0.30 × 0.050 = 0.015 mol
Molar mass = 2.0 / 0.015 ≈ 1.33 × 10² g mol⁻¹.

Q28. 1.8 g of glucose (M = 180 g mol⁻¹) is dissolved in 100 g of water. Calculate the depression in freezing point. (Kf = 1.86 K kg mol⁻¹)

n = 1.8 / 180 = 0.010 mol
m = 0.010 / 0.100 = 0.10 mol kg⁻¹
ΔTf = 1.86 × 0.10 = 0.186 K.

Q29. 0.20 g of an organic solute (M = 200 g mol⁻¹) is dissolved in 20 g of camphor and the freezing point is lowered by 1.40 K. Calculate the molal depression constant (Kf) for camphor.

n = 0.20 / 200 = 0.0010 mol
Solvent mass = 0.020 kg → m = 0.0010 / 0.020 = 0.050 mol kg⁻¹
Kf = ΔTf / m = 1.40 / 0.050 = 28 K kg mol⁻¹.

Q30. If the depression in freezing point of water is 0.372 K, what is the new freezing point? (Pure water freezes at 273.0 K)

New Tf = 273.0 − 0.372 = 272.63 K (≈ 272.6 K).

E. Osmotic Pressure & Molar Mass

Q31. Calculate the osmotic pressure of a 0.10 M aqueous solution at 300 K. (R = 0.0821 L atm K⁻¹ mol⁻¹)

Π = CRT = 0.10 × 0.0821 × 300 ≈ 2.46 atm
Osmotic pressure ≈ 2.46 atm.

Q32. 0.50 g of an unknown solute is dissolved in water to make 50 mL of solution. The osmotic pressure at 300 K is 2.46 atm. Calculate the molar mass of the solute. (R = 0.0821 L atm K⁻¹ mol⁻¹)

Π = CRT → C = Π / (RT) = 2.46 / (0.0821×300) ≈ 0.1 mol L⁻¹
Volume = 0.050 L → n = C×V ≈ 0.1×0.050 = 0.005 mol
Molar mass = 0.50 / 0.005 = 100 g mol⁻¹.

Q33. Calculate the osmotic pressure of a 0.20 M aqueous solution at 298 K. (R = 0.0821 L atm K⁻¹ mol⁻¹)

Π = CRT = 0.20 × 0.0821 × 298 ≈ 4.89 atm
Osmotic pressure ≈ 4.9 atm.

Q34. 1.0 g of urea (M = 60 g mol⁻¹) is dissolved in 250 mL of solution at 300 K. Calculate the osmotic pressure. (R = 0.0821 L atm K⁻¹ mol⁻¹)

n = 1.0 / 60 ≈ 0.0167 mol
C = n / V = 0.0167 / 0.250 ≈ 0.0667 mol L⁻¹
Π = 0.0667 × 0.0821 × 300 ≈ 1.64 atm.

Q35. Calculate the osmotic pressure of 0.10 M NaCl solution at 300 K assuming complete dissociation. (R = 0.0821 L atm K⁻¹ mol⁻¹)

For NaCl (→ Na⁺ + Cl⁻): i ≈ 2
Π = iCRT = 2 × 0.10 × 0.0821 × 300 ≈ 4.93 atm
Osmotic pressure ≈ 4.9 atm.

F. van’t Hoff Factor & Abnormal Molar Mass

Q36. A 0.10 molal solution of KCl shows a freezing point depression of 0.348 K. For a non-electrolyte of same molality, ΔTf would be 0.186 K. Calculate the van’t Hoff factor and the degree of dissociation of KCl. (Kf = 1.86 K kg mol⁻¹)

Theoretical ΔTf (nonelectrolyte) = Kfm = 1.86×0.10 = 0.186 K
i = ΔTf(obs) / ΔTf(calc) = 0.348 / 0.186 ≈ 1.87
For KCl → K⁺ + Cl⁻, i = 1 + α
α ≈ 1.87 − 1 = 0.87 (87% dissociation).

Q37. A solute associates in benzene to form dimers. The observed van’t Hoff factor is 0.75. Calculate the degree of association.

For dimerisation: 2A ⇌ A₂
i = 1 − α/2
0.75 = 1 − α/2 → α/2 = 0.25 → α = 0.50
Degree of association = 50%.

Q38. A 0.30 M glucose solution is isotonic with blood at 37 °C (310 K). Calculate the osmotic pressure of blood. (R = 0.0821 L atm K⁻¹ mol⁻¹)

Π = CRT = 0.30 × 0.0821 × 310 ≈ 7.64 atm
Osmotic pressure ≈ 7.6 atm.

Q39. Calculate the depression in freezing point of a 0.050 molal NaCl solution assuming complete dissociation. (Kf = 1.86 K kg mol⁻¹)

For NaCl, i ≈ 2
ΔTf = iKfm = 2 × 1.86 × 0.050 ≈ 0.186 K.

Q40. Calculate the depression in freezing point of a 0.050 molal CaCl₂ solution assuming complete dissociation. (Kf = 1.86 K kg mol⁻¹)

CaCl₂ → Ca²⁺ + 2Cl⁻ → i ≈ 3
ΔTf = 3 × 1.86 × 0.050 = 0.279 K.

G. Molarity–Molality & Related Applications

Q41. A 1.0 M NaCl solution has density 1.04 g mL⁻¹. Calculate its molality. (M(NaCl) = 58.5 g mol⁻¹)

1 L solution mass = 1.04 g mL⁻¹ × 1000 mL = 1040 g
Moles NaCl = 1.0 mol → mass = 58.5 g
Mass of water = 1040 − 58.5 = 981.5 g = 0.9815 kg
Molality m = 1.0 / 0.9815 ≈ 1.02 mol kg⁻¹.

Q42. A 1.0 molal NaCl solution (1.0 mol per 1.0 kg water) has density 1.04 g mL⁻¹. Calculate its molarity. (M(NaCl) = 58.5 g mol⁻¹)

Mass of water = 1000 g, solute = 58.5 g
Mass of solution = 1058.5 g
Volume = 1058.5 g / (1.04 g mL⁻¹) ≈ 1017.8 mL = 1.018 L
Molarity M = 1.0 / 1.018 ≈ 0.98 mol L⁻¹.

Q43. What elevation in boiling point will be produced in water by a 0.10 molal solution of a non-electrolyte? (Kb = 0.52 K kg mol⁻¹)

ΔTb = Kb m = 0.52 × 0.10 = 0.052 K.

Q44. How many moles of glucose must be dissolved in 0.50 kg of water to raise its boiling point by 0.52 K? (Kb = 0.52 K kg mol⁻¹)

m = ΔTb / Kb = 0.52 / 0.52 = 1.0 mol kg⁻¹
For 0.50 kg water: n = 1.0 × 0.50 = 0.50 mol.

Q45. What mass of NaCl (M = 58.5 g mol⁻¹) should be dissolved in 200 g of water to lower its freezing point by 0.372 K? Assume complete dissociation and Kf = 1.86 K kg mol⁻¹.

For NaCl, i ≈ 2
ΔTf = i Kf m → m = ΔTf / (iKf) = 0.372 / (2×1.86) = 0.10 mol kg⁻¹
Solvent mass = 0.200 kg → n = 0.10 × 0.200 = 0.020 mol
Mass = 0.020 × 58.5 = 1.17 g.

Q46. A 0.10 M electrolyte solution at 300 K shows an osmotic pressure of 4.92 atm. For a non-electrolyte of same concentration, Π would be 2.46 atm. Calculate the van’t Hoff factor.

i = Π(obs) / Π(calc for nonelectrolyte) = 4.92 / 2.46 = 2.0 (≈ two ions per formula unit).

Q47. Calculate the depression in freezing point of a 0.20 molal BaCl₂ solution assuming complete dissociation. (Kf = 1.86 K kg mol⁻¹)

BaCl₂ → Ba²⁺ + 2Cl⁻ → i ≈ 3
ΔTf = iKfm = 3 × 1.86 × 0.20 = 1.116 K.

Q48. 5.0 g of urea (M = 60 g mol⁻¹) is dissolved in 100 g of water. Calculate the relative lowering of vapour pressure.

n(solute) = 5.0 / 60 ≈ 0.0833 mol
n(solvent) = 100 / 18 ≈ 5.56 mol
X(solute) = 0.0833 / (0.0833 + 5.56) ≈ 0.0148
Relative lowering Δp/p⁰ = X(solute) ≈ 1.48 × 10⁻².

Q49. Using data from Q48, if vapour pressure of pure water at that temperature is 23.8 mm Hg, calculate the vapour pressure of the solution.

p = p⁰ (1 − Δp/p⁰) = 23.8 × (1 − 0.0148) ≈ 23.8 × 0.9852 ≈ 23.45 mm Hg
Vapour pressure ≈ 23.4 mm Hg.

Q50. How much water should be added to 200 mL of 1.5 M NaCl solution to obtain a 0.50 M solution?

M₁V₁ = M₂V₂ → 1.5 × 200 = 0.50 × V₂
V₂ = (1.5 × 200) / 0.50 = 600 mL
Final volume = 600 mL; water added = 600 − 200 = 400 mL.
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