NaCl is doped with SrCl2 at 1.2x10^-2 mol%. Each Sr^2+ replaces two Na+ ions' worth of charge by creating one cationic vacancy. Using N_A=6.022x10^23 mol^-1, find the number of cationic vacancies in 2.00 mol of NaCl in units of 10^20 (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ss099

NaCl is doped with $SrCl X_{2}$ at $1.2 \times 1 0^{- 2}$ mol%. Each $Sr X^{2 +}$ replaces two $Na X^{+}$ ions' worth of charge by creating one cationic vacancy. Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the number of cationic vacancies in $2.00$ mol of NaCl in units of $1 0^{20}$ (two decimals).

Numerical answer type, answer in x 10^20, tolerance 0.02.

Mole fraction of

SrCl X_{2}

= 1.2 \times 1 0^{- 2} % = 1.2 \times 1 0^{- 4}

. Each

Sr X^{2 +}

gives one vacancy, so vacancies

= 2.00 \times 1.2 \times 1 0^{- 4} \times 6.022 \times 1 0^{23} = 1.4453 \times 1 0^{20} \approx 1.45 \times 1 0^{20}