One mole of AgCl is doped with 2.0x10^-4 mol of CdCl2, each Cd^2+ replacing an Ag+ ion. With N_A = 6.022x10^23 mol^-1, the number of cation vacancies created is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ss201

One mole of $AgCl$ is doped with $2.0 \times 1 0^{- 4} mol$ of $CdCl X_{2}$ , each $Cd X^{2 +}$ replacing an $Ag X^{+}$ ion. With $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , the number of cation vacancies created is

Each

Cd X^{2 +}

replaces one

Ag X^{+}

and creates one vacancy for charge balance. Vacancies

= 2.0 \times 1 0^{- 4} \times 6.022 \times 1 0^{23} = 1.2044 e + 20

, i.e.

1.2044 \times 1 0^{20}

. Doubling this would wrongly assume two vacancies per

Cd X^{2 +}