A sample of 5.85 g of NaCl (M=58.5 g mol^-1) has a fraction 2.5x10^-8 of its cation sites vacant because of Schottky defects. Using N_A=6.022x10^23 mol^-1, find the number of cation vacancies in units of 10^14 (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3ss100

A sample of $5.85 g$ of NaCl ( $M = 58.5 g mo l^{- 1}$ ) has a fraction $2.5 \times 1 0^{- 8}$ of its cation sites vacant because of Schottky defects. Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the number of cation vacancies in units of $1 0^{14}$ (nearest integer).

Numerical answer type, answer in x 10^14, tolerance 0.5.

Formula units

= \frac{5.85}{58.5} \times 6.022 \times 1 0^{23} = 6.022 \times 1 0^{22}

, equal to the number of cation sites. Vacancies

= 6.022 \times 1 0^{22} \times 2.5 \times 1 0^{- 8} = 15.055 \times 1 0^{14} \approx 15 \times 1 0^{14}