MCQ Solution: Stoichiometry
The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to, an excess of potassium iodide solution liberating iodine according to the following equation 2Cu2 (aq) + 4I– (aq) ⇌2CuI(s) + I2(aq) The iodine liberated required 24.5cm3 of a 0.100 mole dm-3 solution of sodium thiosulphate 2S2O32- (aq) + I2(aq) → 2I– (aq) + S4O62- (aq) the percentage of copper, by mass in the copper(ll) salt is. [Atomic mass of copper = 63.5]
1) 64.2
2) 51.0
3) 48.4
4) 25.5
Explanation:
From given reactions
mmoles of hypo = mmoles of iodine × 2
= mmoles of Cu2+ ions
= 24.5 × 0.1 mmoles
So mass of copper = 24.5 × 0.1 × 10–3 × 63.5 gm
So % of copper =
≃ 51.0%