CaF2 (molar mass 78 g mol^-1) crystallises with an fcc lattice of Ca^2+ and cell edge 546 pm. Given N_A = 6.022x10^23 mol^-1, calculate its density in g cm^-3 (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ss176

$CaF X_{2}$ (molar mass $78 g mo l^{- 1}$ ) crystallises with an fcc lattice of $Ca X^{2 +}$ and cell edge $546 pm$ . Given $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , calculate its density in $g c m^{- 3}$ (two decimals).

Numerical answer type, answer in g/cm^3, tolerance 0.01.

The cell holds 4

Ca X^{2 +}

and 8

F X^{-}

, i.e.

Z = 4

formula units.

ρ = \frac{4 \times 78}{6.022 \times 1 0 ^{23} \times ( 5.46 \times 1 0 ^{- 8} ) ^{3}} = 3.18 g c m^{- 3}