Board-style MCQs and past-paper questions from Karachi coaching notes. Tap an option to check yourself instantly. Solved numericals are at the bottom.
Multiple-choice questions
Solved numericals & self-assessment (past papers)
Graphite — atoms & moles in 360 g of carbon
moles = mass / molar mass = 360 / 12 = 30 mol
no. of atoms = moles × Nₐ = 30 × 6.02 × 10²³
= 1.807 × 10²⁵ atoms (30 mol)
Ammonium nitrate decomposition — mass of water & volume of N₂O
NH₄NO₃ → N₂O + 2H₂O (80 g NH₄NO₃ per mole)
Mass of H₂O: 80 g NH₄NO₃ → 36 g H₂O, so 200 g → (36/80) × 200 = 90 g H₂O
Volume of N₂O: 80 g NH₄NO₃ → 22.4 dm³ N₂O, so 200 g → (22.4/80) × 200 = 56 dm³ N₂O
Magnesium + oxygen — percentage yield of MgO
2Mg + O₂ → 2MgO. 4 g Mg gives 4.24 g MgO (practical).
moles Mg = 4 / 24 = 0.166 mol → 0.166 mol MgO
theoretical yield = 0.166 × 40 = 6.64 g
% yield = (4.24 / 6.64) × 100 = 63.8 %
KClO₃ heating — mass of KCl & O₂ from 49 g
2KClO₃ → 2KCl + 3O₂
moles KClO₃ = 49 / 122.5 = 0.4 mol
O₂ = (3/2)(0.4) × 32 = 19.2 g
KCl = 0.4 × 149 = 29.8 g
Na₂S₂O₃ preparation — limiting reactant & leftover Na₂S
2Na₂S + Na₂CO₃ + 4SO₂ → 3Na₂S₂O₃ + CO₂, 200 g of each.
SO₂ is the limiting reactant (basic medium; Na₂S left over).
moles SO₂ = 200 / 64 = 3.125 mol → (3/4)(3.125) = 2.343 mol Na₂S₂O₃
mass Na₂S₂O₃ = 2.343 × 158 = 370.2 g
Na₂S used = 121.84 g → remaining = 200 − 121.84 = 78.16 g