Grade 12th passPhysical Chemistry
If NaCl is dropped with 10^-3mole% of SrCl2.Calculate the number of cationic vacancies.
1 Answer
Umakant biswal
Approved Tutor Answer9 Years ago@ 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 ../
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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