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Questions  >>  JEEMAIN and NEET  >>  NEET PAST PAPERS  >>  2007
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Q)

If $NaCl$ is doped with $10^{-4}$ mole % of $SrCl_2$, the concentrate of cation vacancies will be : $(N_A = 6.02 \times 10^{23}) mol^{-1}$


( A ) $6.02 \times 10^{14} mol^{-1}$
( B ) $6.02 \times 10^{16} mol^{-1}$
( C ) $6.02 \times 10^{15} mol^{-1}$
( D ) $6.02 \times 10^{17} mol^{-1}$

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