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Karnataka Board PUCPUC Science 2nd PUC Class 12

If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies? - Chemistry

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Question

If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies?

Numerical
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Solution

When NaCl is doped with SrCl2, each additional Sr2+ ion eliminates two Na+ ions, creating two vacancies, one occupied by Sr2+ and the other unoccupied. Thus, adding one Sr2+ ion results in one cation vacancy.

Doping of NaCl with 1 × 10−3 rnol % of SrCl2 means that 1 × 10−3 moles of NaCl are added to 100 moles of NaCl. 

∴ 1 mole of NaCl contains SrCl2 = `(1 xx 10^-3)/100`

= 1 × 10−5 moles

= 1 × 10−5 × 6.022 × 1023 molecules

= 6.022 × 1018 molecules

∴ No. of Sr2+ ions present in one mole of NaCl = No. of cation vacancies in one mole of NaCl

= 6.022 × 1018

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Chapter 1: Solid State - 'NCERT TEXT-BOOK' Exercises [Page 55]

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Nootan Chemistry Part 1 and 2 [English] Class 12 ISC
Chapter 1 Solid State
'NCERT TEXT-BOOK' Exercises | Q 1.25 | Page 55
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