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If NaCl is dropped with 10^-3mole% of SrCl2.Calculate the number of cationic vacancies.
@ vatsala concentration of srcl2 = 10 ^ -2 mole percent assume we have 100 moles of nacl solution . then 10 ^ -3 moles of srcl is doped in it . 1 mole of nacl is doped with 10 ^ -3 / 100 moles of srcl2 when a na+ is replaced with sr 2+ , 1 cationic vacancies are produced . so, total no of cationic vacancies produced by 10 ^ -5 moles of srcl2 = 10^-5 * 6.023 * 10 ^ 23 = 6.023 * 10^18 per mole . HOPE IT CLEARS YOUR DOUBT ALL THE BEST ../
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