Question

If NaCl is doped with 10^-4 mol % SrCl₂ then concentration of cation vacancies will be

a. 6.022 x 10¹⁴ mol^-1

b. 6.022 x 10¹⁵ mol^-1 

c. 6.022 x 10⁶ mol^-1

d. 6.022 x 10¹⁷ mol^-1

Answer 1

d. 6.022 x 10¹⁷ mol^-1

NaCl is doped with 10 mol % of SrCl₂

⇒ 100 mol of NaCl is doped with 10^-4 mol of SrCl₂

∴ 1 mol of NaCl is doped with \(\frac{10^{-4}}{100}\) mol of SrCl₂

= 10^-6 mol of SrCl₂

Number of cation vacancies produced by each Sr²⁺ = 1

∴ Concentration of cation vacancies produced by 10^-6 mol of Sr²⁺ ions = 10^-6 x 6.022 x 10²³

= 6.022 x 10¹⁷ mol^-1