Solutions: Class 12 Chemistry Q&A
7 free answers on Solutions (Chemistry) for Class 12 students, across CBSE, ICSE and State boards. Includes questions such as “P°(hexane) = 408 Torr, P°(heptane) = 141 Torr. x(hexane) = 0.300. Find Y₆ and Y₇.”, “O₂ at 293 K, partial pressure = 0.98 bar, KH = 34 kbar. Find solubility in g/L.”, “Find density (g/mL) of 3.60 M H₂SO₄ that is 29% by mass. (Mᵣ = 98 g/mol)”.
Question & Answer
Questions
- ChemistryClass 12CBSE
P°(hexane) = 408 Torr, P°(heptane) = 141 Torr. x(hexane) = 0.300. Find Y₆ and Y₇.
Step-1: Partial pressures:P(hexane) = 0.300 × 408 = 122.4 TorrP(heptane) = 0.700 × 141 = 98.7 TorrStep-2: Total pressure:P_total = 122.4 + 98.7 = 221.1 TorrStep-3: Vapour mole fractions:Y₆ (hexane) = 122.4 / 221.1 ≈ 0.554Y₇ (heptane) = 98.7 / 221.1 ≈ 0.446Answer: Y₆ ≈ 0.554, Y₇ ≈ 0.446
- ChemistryClass 12CBSE
P°A = 7×10³ Pa, P°B = 12×10³ Pa at 350 K. Liquid contains 40 mol% A. Find vapour composition.
Step-1: Partial pressures (Raoult’s Law):PA = 0.40 × 7000 = 2800 PaPB = 0.60 × 12000 = 7200 PaStep-2: Total pressure:P_total = 2800 + 7200 = 10000 PaStep-3: Vapour mole fractions (Dalton’s Law):yA = 2800 / 10000 = 0.28yB = 7200 / 10000 = 0.72Answer: Mole fraction of A in vapour = 0.28, B in vapour = 0.72The more volatile component (B, higher P°) is enriched in the
- ChemistryClass 12CBSE
O₂ at 293 K, partial pressure = 0.98 bar, KH = 34 kbar. Find solubility in g/L.
Step-1: Mole fraction:x(O₂) = 0.98 / 34000 = 2.882×10⁻⁵Step-2: Moles of O₂ in 1 L:n(O₂) ≈ 2.882×10⁻⁵ × 55.5 = 1.60×10⁻³ molStep-3: Mass (Mᵣ of O₂ = 32 g/mol):mass = 1.60×10⁻³ × 32 = 0.0512 gAnswer: Solubility of O₂ ≈ 0.0512 g/LO₂ has a smaller KH than N₂ (34 vs 76.48 kbar) meaning O₂ is more soluble, which is why aquatic life can survive using dissolved O₂.
- ChemistryClass 12CBSE
Find density (g/mL) of 3.60 M H₂SO₄ that is 29% by mass. (Mᵣ = 98 g/mol)
Formula linking molarity, density, and mass percent:M = (10 × d × w%) / Molar massSubstituting values:3.60 = (10 × d × 29) / 98d = (3.60 × 98) / (10 × 29) = 352.8 / 290 = 1.216 g/mLAnswer: Density = 1.216 g/mLNote: Memorise M = (10×d×w%)/Mᵣ. Always keep w% as a plain number (29, not 0.29) and density in g/mL.
- ChemistryClass 12CBSE
N₂ gas at 293 K, partial pressure = 0.987 bar, KH = 76.48 kbar. Find millimoles dissolved in 1 L water.
Step 1: Apply Henry’s Law (p = KH × x):x(N₂) = 0.987 / 76480 = 1.290×10⁻⁵Step 2: Moles of water in 1 L = 1000/18 = 55.5 molStep 3: Moles of N₂ (since x << 1):n(N₂) ≈ x × n(H₂O) = 1.290×10⁻⁵ × 55.5 = 7.16×10⁻⁴ mol = 0.716 mmolAnswer: ≈0.716 millimoles of N₂ dissolveKey Point: Henry’s Law: gas solubility ∝ partial pressure. The large KH for N₂ means it is only sparingly soluble a
- ChemistryClass 12CBSE
50 g of ethylene glycol (CH₂OH)₂ dissolved in 200 g water; cooled to −9.3°C. Find mass of ice that separates. (Kf = 1.86 K·kg/mol)
Molar mass of ethylene glycol (C₂H₆O₂) = 62 g/molStep 1 — Moles of solute:n = 50 / 62 = 0.8065 molStep 2 — Freezing point depression formula: ΔTf = Kf × m9.3 = 1.86 × (0.8065 / W_solvent_kg)Step 3 — Mass of liquid water remaining:W_solvent = (1.86 × 0.8065) / 9.3 = 0.1613 kg = 161.3 gStep 4 — Ice separated = total water − liquid water:Ice = 200 − 161.3 = 38.7 gAnswer: ≈38.71 g of ice separates outWhen a sol
- ChemistryClass 12All
What is Delta T = K_f × m?
This is the freezing point depression formula: ΔT = freezing point lowering (°C) K_f = cryoscopic constant (specific to solvent) m = molality of solution (mol/kg) For boiling point elevation: ΔT = K_b × m These colligative properties depend on solute concentration, not identity.