If NaCl is doped with 10-5 mole percent AlCl3, then the number of cation vacancies per mole will be (1) 6.023 x 1023 (2) 6.023 x 1018 (3) 6.023 x 1016 (4) 1.2046 x 1017 Share with your friends Share 4 Priya answered this Full explains -24 View Full Answer Kylo Ren answered this 10^-5 mole percent of alcl3=10^-5/100 mole al+3= 10^-7 mole therefore, cationic vacancies per mole=10^-7 *Na=6.023*10^16 -5
