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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Solid State
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If NaCl is doped with $10^{–4}$ mol% of $SrCl_2$, the concentration of cation vacancies will be $(N_A = 6.02 \times 10^{23} mol^{–1}$)

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$6.02 \times 10^{17} mol^{–1 }$
Hence (C) is the correct answer.
answered Jun 3, 2014 by sreemathi.v
 
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