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If NaCl is doped with 10 raise to the power -4 mol % of SrCl2, then what will be the concentration of cation vacancies? given-Avogado's no.- 6.02 * 10 raise to the power 23

plzzzzzz somebody give the solution as soon as possible

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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
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hey ankita are u a bio student 

yaar actually i am a bio student and i have not opted maths  but i know ica n score 90+ in boards but yaar i have a little bit confusion that in stephan's if i want to pursue bsc chemistry

am i eligible for it on the behalf of PCB???????????

PLEASE HELP ME OUT?????????????

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ya ya definitely rashika u r eligible for it ok,,,,see any person who opts for medical wants to be a doctor u carry on ur studies according to that ok and then see will u be able to crack the entrance ok if u can then what else? and if u can't then u can very well opt for bse or msc .

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thank u very much friend

friend if u r a bio student then u can explain me the page no-84 of biology???

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okkkkkkkkkkk,,,,,i will definitely send u a satisfactory answer very soon........ley me first check out tc rashika 

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rashika,,,if i am not wrong then you are talking about 84th page in 12th NCERT book? just confirm it once as soon as possible ok so that i may help u accordingly if i would be elligible for that

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Number of moles of cationic vacancies =100Mole Of nacl-10^-4No. Of moles of cationic vacancies per mole of nacl-10^-6Total no. Of cationic vacancies -6.02*10^17
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