If Nacl is doped with 10-3 mole % of Srcl2 , what is the concentration of cation vacancies?
NaCl is doped with 10-3 mole % of SrCl2.This means, 100 moles of NaCl are doped with 10-3 moles of SrCl2.
Therefore, 1mole of NaCl is doped with 10-3 /100 = 10-5 moles of SrCl2 or 10-5 moles of Sr2+.
One Sr2+ ion create one cation vacancy.
Therefore, no. of cation vacancies created by 10-5 moles Sr2+
= 10-5 moles/mole of NaCl
= 10-5 x 6.022 x 1023 / mole of NaCl
= 6.022 x 1018/ mole of NaCl
It is an NCERT question, Q.25
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 withmol of SrCl2
= 10−5 mol of SrCl2
Cation vacancies produced by one Sr2+ ion = 1
Hence, the concentration of cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.