Question

A certain sample of cuprous sulphide is found to have the composition $C{u}_{1.95}{S}_{1.00}$ because of incorporation of $C{u}^{2+}$ and $C{u}^{+}$ ions in the crystal then ratio of $C{u}^{2+}$ and $C{u}^{+}$ ion is:

• A. 0.08 and 1.00
• B. 1: 38
• C. 1 : 24
• D. 1.00 : 1.00

Answer 1

Answer: (B)

1: 38

Cuprous sulphide has molecular formula $C{u}_{2}S$. One mole of $C{u}^{+2}$ replaces two moles of $C{u}^{+}$.

Moles of $C{u}^{+2}$ present in the sample $=2-1.95=0.05$

Percentage of $C{u}^{+2}=\frac{0.05}{1.95}\times 100=2.56$

Percentage of $C{u}^{+}=100-2.56=97.44$

Ratio of $C{u}^{+2}$ and $C{u}^{+}=\frac{2.56}{97.44}=\frac{1}{38}=1:38$