In a rock-salt type solid MX the anion radius is 190 pm and the cation just fits the octahedral void, so r_+/r_- = sqrt2 - 1 (take sqrt2=1.414). Cations and anions touch along the cell edge. If the molar mass of MX is 60 g mol^-1 and N_A = 6.022x10^23 mol^-1, find the density in g cm^-3 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss179

In a rock-salt type solid $MX$ the anion radius is $190 pm$ and the cation just fits the octahedral void, so $r_{+} / r_{-} = 2 - 1$ (take $2 = 1.414$ ). Cations and anions touch along the cell edge. If the molar mass of $MX$ is $60 g mo l^{- 1}$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the density in $g c m^{- 3}$ (two decimals).

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

r_{+} = 0.414 \times 190 = 78.66 pm

. Contact along the edge gives

a = 2 (r_{+} + r_{-}) = 2 (78.66 + 190) = 537.32 pm = 5.3732 \times 1 0^{- 8} cm

. With

Z = 4

ρ = \frac{4 \times 60}{6.022 \times 1 0 ^{23} a ^{3}} = 2.57 g c m^{- 3}