A non-stoichiometric compound Cu1 . 8S is formed due to the incorporation of Cu2 + ions in the lattice. What is the mole percentage of Cu2 + present in the compound?

Question

A non-stoichiometric compound Cu1 . 8S is formed due to the incorporation of Cu2 + ions in the lattice. What is the mole percentage of Cu2 + present in the compound? (A) 88.88 (B) 89.8 (C) 63.5 (D) 11.11

Tardigrade Admin · Accepted Answer

In one mole, Let there be Cu +2=x moles ∴ Cu +=1.8- x moles ∵ 2 x +(1.8- x ) 1=+2 x =0.2 % Cu +2=(0.2/1.8) × 100=11.11 %

Q. A non-stoichiometric compound $C u_{1.8} S$ is formed due to the incorporation of $C u^{2 +}$ ions in the lattice. What is the mole percentage of $C u^{2 +}$ present in the compound?

$88.88$

$89.8$

$63.5$

$11.11$

In one mole, Let there be $C u^{+ 2} = x$ moles
$∴ C u^{+} = 1.8 - x$ moles
$∵ 2 x + (1.8 - x) 1 = + 2$
$x = 0.2$
$% C u^{+ 2} = \frac{0.2}{1.8} \times 100 = 11.11%$

Q. A non-stoichiometric compound Cu1.8​S is formed due to the incorporation of Cu2+ ions in the lattice. What is the mole percentage of Cu2+ present in the compound?

88.88

89.8

63.5

11.11