KBr is doped with $10^{-5}$ mole percent of $SrBr_2$. The number of cationic vacancies in 1 g of KBr

Question

KBr is doped with $10^{-5}$ mole percent of $SrBr_2$. The number of cationic vacancies in 1 g of KBr crystal is $X 	imes 10^{14}$ (Round off to the Nearest Integer) [Atomic mass: K: 39.1 u, Br: 79.9 u]

Professor PC Thomas Classes · Answer

Molar mass KBr = 119 g/mol. Moles of KBr in 1 g = 1/119. Cationic vacancies = moles of $Sr^{2+}$ added = $10^{-5} / 100 = 10^{-7}$ per mole of KBr. Total vacancies = $10^{-7} 	imes (1/119) 	imes 6.022 	imes 10^{23} \approx 5.06 	imes 10^{14}$. So X = 5.