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JEE Advanced levelNumerical answerDifficulty 3 of 3ss040

A sample of NaCl (cubic, edge $564 pm$ , 4 formula units per ideal cell, $M = 58.5 g mo l^{- 1}$ ) has measured density 2.1544 $g c m^{- 3}$ because of Schottky defects, the cell edge being unchanged. Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the percentage of $Na X^{+}$ - $Cl X^{-}$ pair sites left vacant (two decimals).

Numerical answer type, answer in %, tolerance 0.03.

Answer: 0.53 %

Ideal density

= \frac{4 \times 58.5}{6.022 \times 1 0 ^{23} \times ( 5.64 \times 1 0 ^{- 8} ) ^{3}} = 2.1659 g c m^{- 3}

. With the same cell volume, density is proportional to the number of ion pairs present. Fraction vacant

= 1 - \frac{2.1544}{2.1659} = 0.00531

, i.e. 0.53%.

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