Question

lf nacl is dopped with0.001 moleof bacl2 then the concentration of cationic vacancies is​

Answer 1

hello!

_____

Given Conetration of SrCl2 = 10−3 mol%

Concentration is in percentage so that take total 100 mol of solution

Number of moles of NaCl = 100 – moles of SrCl2

Moles of SrCl2is very negligible as compare to

total moles so

Number of moles of NaCl = 100

1 mol of NaCl is dipped with = 10−3/100 moles of SrCl2

1 mol of NaCl is dipped with = 10−3/100 moles of SrCl2 = 10–5 mol of SrCl2

So cation vacancies per mole of NaCl =10–5 mol

So cation vacancies per mole of NaCl =10–5 mol1 mol = 6.022 x1023 particles

So

So cation vacancies per mole of NaCl = 10–5 x 6.022 x1023

So cation vacancies per mole of NaCl = 10–5 x 6.022 x1023 = 6.02 x1018

So that, the concentration of cation vacancies created by SrCl2is 6.022 × 108 per mol of NaCl.