If NaCl is doped with 10-3 mole % of SrCl2, then number of cationic vacancies is

Question

If NaCl is doped with 10-3 mole % of SrCl2, then number of cationic vacancies is (A) 6.02 × 10-18 mol-1 (B) 10-5 mol-1 (C) 6.02 × 1020 mol-1 (D) 6.02 × 1018 mol-1

Tardigrade Admin · Accepted Answer

Due to the addition of SrCl2, each Sr2+ ion replaces two Na+ ions, but occupies one Na+ lattice point. Thus, this exchange of Na+ ion by Sr2+ ion makes one cationic vacancy. SrCl2 doped = 10-3 mol per 100 mol = 10-5 mol per 1 mol ∴ Cation vacancies = 10-5 mol per 1 mol = 10-5 × N0 mol-1 = 10-5 × 6.02 × 1023 Total = 6.02 × 1018 cationic vacancies mol-1

Q. If $N a Cl$ is doped with $1 0^{- 3}$ mole $%$ of $S r C l_{2}$ , then number of cationic vacancies is

$6.02 \times 1 0^{- 18} m o l^{- 1}$

$1 0^{- 5} m o l^{- 1}$

$6.02 \times 1 0^{20} m o l^{- 1}$

$6.02 \times 1 0^{18} m o l^{- 1}$

Q. If NaCl is doped with 10−3 mole % of SrCl2​, then number of cationic vacancies is

6.02×10−18mol−1

10−5mol−1

6.02×1020mol−1

6.02×1018mol−1

Q. If $N a Cl$ is doped with $1 0^{- 3}$ mole $%$ of $S r C l_{2}$ , then number of cationic vacancies is

$6.02 \times 1 0^{- 18} m o l^{- 1}$

$1 0^{- 5} m o l^{- 1}$

$6.02 \times 1 0^{20} m o l^{- 1}$

$6.02 \times 1 0^{18} m o l^{- 1}$