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  • KBr is doped with 10-5 mole percent of SrBr2 . The number of cationic vacancies in 1 g of KBr crystal is ________.1014 (Round off to the Nearest Integer). [Atomic Mass : K : 39.1 u, Br : 79.9 u, NA = 6

KBr is doped with 10-5 mole percent of SrBr2 . The number of cationic vacancies in 1 g of KBr crystal is ________.1014 (Round off to the Nearest Integer). [Atomic Mass : K : 39.1 u, Br : 79.9 u, NA = 6.023 x 1023]
 

Answers (1)

best_answer

The molar mass of KBr = 119 g/mol.

For every Sr+2 ion, 1 cationic vacancy is created.

Hence, no. of Sr+2 ion = Number of cationic vacancies.

GIven,

KBr is doped with 10–5 mole percent of SrBr2 .

So, 1 mole KBr have \frac{10^{-5}}{100} \text { moles } \mathrm{SrBr}_{2} or  10-7 moles cation vacancies.

In 1 mol of KBr 10-7 moles, cation vacancies are present

Now, 

In 119 g/mol  of KBr , 10-7 moles of cation vacancies are present.

In 1 g/mol of KBr , \frac{10^{-7}}{119} moles cation vacancies are present.

Now,  in 1 gram of KBr only 

\text {No. of cationic vacancy }=\frac{10^{-7}}{119} \times \mathrm{N}_{\mathrm{A}}

=\frac{1}{119} \times 10^{-7} \times 6.022 \times 10^{23}

=5 \times 10^{-2} \times 10^{-7} \times 10^{23}

=5 \times 10^{14}

Ans = 5

Posted by

Kuldeep Maurya

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