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) $\stackrel{}{⇌}$ 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) $\underset{}{\overset{}{\to }}$ 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

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(2) 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 = $\frac{24.5\text{\hspace{0.17em}}×\text{\hspace{0.17em}}0.1\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{-3}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}63.5}{0.305}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}100%\text{\hspace{0.17em}}\simeq \text{\hspace{0.17em}}51.0\text{\hspace{0.17em}}%$

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