Q.13 If NaCl is doped with 10-4 % SrCl2, the % of cationic vacancy is, ?
Asked by Pulkit | 22nd May, 2012, 10:12: AM
On doping NaCl by SrCl2, one Sr 2+ ion replaces two Na+ ion.
So, no. of moles of cation vacancy in 100ml NaCl = 10-4
no of moles of cation vacancy in 1 mole NaCl = 10-4/100 = 10-6
Thus, total cation vacancy = 10-6 x N0
= 10-6 x 6.022 x 1023 = 6.022 x 1017
Answered by | 23rd May, 2012, 01:00: PM
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