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If NaCl is doped with 10^-3 mol % of SrCl2, what is the concentration of cation vacancies?

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If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?

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It is given that NaCl is doped with 10-3 mol% of SrCl2.

This means that 100 mol of NaCl is doped with 10-3 mol of SrCl2.

Therefore, 1 mol of NaCl is doped with \(\frac{10^{-3}}{100}\) mol of SrCl2

= 10-3 mol% of SrCl2

Cation vacancies produced by one Sr2+ ion = 1

∴ Concentration of the cation vacancies produced by 10-5 mol of Sr2+ ion = 10-5 x 6.022 x 1023

= 6.022 x 1018mol-1

Hence, the concentration of cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.

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