KBr is doped with 10^{-5} mole percent of SrB | vedclass

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(D) Molar mass of $KBr = 39.1 + 79.9 = 119 \ g/mol$.$1 \ mole$ of $KBr$ is doped with $\frac{10^{-5}}{100} = 10^{-7} \ moles$ of $SrBr_2$.Since $1 \ S

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(D) Molar mass of $KBr = 39.1 + 79.9 = 119 \ g/mol$.$1 \ mole$ of $KBr$ is doped with $\frac{10^{-5}}{100} = 10^{-7} \ moles$ of $SrBr_2$.Since $1 \ Sr^{2+}$ ion replaces $2 \ K^+$ ions to maintain electrical neutrality,it creates $1$ cationic vacancy.Therefore,$1 \ mole$ of $KBr$ contains $10^{-7} \ moles$ of cationic vacancies.Number of cationic vacancies in $1 \ g$ of $KBr = \frac{10^{-7} 	imes N_A}{	ext{Molar mass of } KBr} = \frac{10^{-7} 	imes 6.023 	imes 10^{23}}{119}$.$= \frac{6.023 	imes 10^{16}}{119} \approx 0.0506 	imes 10^{16} = 5.06 	imes 10^{14}$.Rounding to the nearest integer,we get $5 	imes 10^{14}$.

$KBr$ is doped with $10^{-5}$ mole percent of $SrBr_2$. The number of cationic vacancies in $1 \ g$ of $KBr$ crystal is ........ $\times 10^{14}$. (Round off to the Nearest Integer).
[Atomic Mass : $K = 39.1 \ u$,$Br = 79.9 \ u$; $N_A = 6.023 \times 10^{23} \ mol^{-1}$]

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$KBr$ is doped with $10^{-5}$ mole percent of $SrBr_2$. The number of cationic vacancies in $1 \ g$ of $KBr$ crystal is ........ $\times 10^{14}$. (Round off to the Nearest Integer).[Atomic Mass : $K = 39.1 \ u$,$Br = 79.9 \ u$; $N_A = 6.023 \times 10^{23} \ mol^{-1}$]