A certain sample of cuprous sulphide is found to have composition Cu_1.8S, because of incorporation of cu²⁺ ions in the lattice. What is the mole % of cu²⁺ in total copper content in this crystal?

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88.88%

11.11%`

99.8%

None of these

answer is B.

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Detailed Solution

Two Cu⁺ replaced by one Cu²⁺
Total loss of Cu = 0.2 mole
So, mole of Cu²⁺ present in Cu_1.8S = 0.2
$â´$ % of Cu ²⁺ in total copper content
$= \frac{0 .2}{1 .8} Ã 100 = 11 .11$

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