Question

# If NaCl is doped with $$10^{-3}$$ mol percent of $$SrCl_2$$, what is the number of cation vacancies?

A
3.01×1018 mol1
B
6.02×1018 mol1
C
1.2×1019 mol1
D
18×1018 mol1

Solution

## The correct option is C $$6.02\, \times\, 10^{18}$$ mo$$l^{-1}$$If NaCl is doped with $$10^{-3}$$ mol percent of $$SrCl_2$$, what is the number of cation vacancies is $$6.02\, \times\, 10^{18}$$ mo$$l^{-1}$$.$$100$$ moles of $$NaCl$$ contains $$10^{-3}$$ moles of $$SrCl_2$$1 mole of NaCl will contain $$\displaystyle 6.02\, \times\, 10^{23} \times \frac {10^{-3}}{100}=6.02\, \times\, 10^{18}$$ molecules of $$SrCl_2$$.1 molecule of $$SrCl_2$$ will lead to one cation vacancy.$$6.02\, \times\, 10^{18}$$ molecules of $$SrCl_2$$ will give $$6.02\, \times\, 10^{18}$$ cation vacanciesChemistry

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