SATHEE: The Solid State - Result Question 36

The Solid State - Result Question 36

####36. If $N a C l$ is doped with $10^{- 4} m o l$ of $S r C l_{2}$ , the concentration of cation vacancies will be

$(N_{A} = 6.02 \times 10^{23} m o l^{- 1})$

[2007]

(a) $6.02 \times 10^{16} m o l^{- 1}$

(b) $6.02 \times 10^{17} m o l^{- 1}$

(c) $6.02 \times 10^{14} m o l^{- 1}$

(d) $6.02 \times 10^{15} m o l^{- 1}$

Show Answer

Solution:

(b) For each $S r^{2 +}$ ion added, one $N a^{+}$ ion is removed to maintain the electrical neutrality. Hence concentration of cation vacancies $=$ mole of $S r C l_{2}$ added $= 10^{- 4} m o l e$

$= \frac{10^{- 4}}{100} \times 6.023 \times 10^{23} = 6.023 \times 10^{17}$

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