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Chemistry Question on The solid state

If NaClNaCl is doped with 10−4 mol %10^{-4}\, mol \, \% of SrCl2,SrCl_2, the concentration of cation vacancies will be (NA=6.023×1023 mol−1)(N_A = 6 . 0 2 3 \times 10^{23}\,mol^{-1})

A

6.023×1015 mol−16 . 0 2 3 \times 10^{15}\,mol^{-1}

B

6.023×1016 mol−16 . 0 2 3 \times 10^{16}\,mol^{-1}

C

6.023×1017 mol−16 . 0 2 3 \times 10^{17}\,mol^{-1}

D

6.023×1014 mol−16 . 0 2 3 \times 10^{14}\,mol^{-1}

Answer

6.023×1017 mol−16 . 0 2 3 \times 10^{17}\,mol^{-1}

Explanation

Solution

Doping of NaClNaCl with 10−4mol%of SrCl210^{-4} mol \% of \, SrCl_2 means, 100100 moles of NaClNaCl are doped with 10−4mol of SrCl210^{-4} mol\, of\, SrCl_2 ∴\therefore 11 mol of NaClNaCl is doped with                          SrCl2=10−4100=10−6mole\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, SrCl_2=\frac{10^{-4}}{100}=10^{-6} mole As each Sr2+Sr^{2+} ion introduces one cation vacancy. ∴\therefore Concentration of cation vacancies                       =10−6\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =10^{-6} mol of NaClNaCl                       =10−6×6.023×1023mol−1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =10^{-6}\times6.023 \times10^{23}mol^{-1}                       =6.023×1017mol−1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =6.023\times10^{17} mol^{-1}