The compound M2O (molar mass= 200 g mol-1 ) has anti-fluorite closest unit cell. If the radius of M+ is 79.55 pm, the density of M2Owould be

The compound M₂O (molar mass= 200 g mol^-1 ) has anti-fluorite closest unit cell. If the radius of M+ is 79.55 pm, the density of M₂Owould be

A
0.67 g mol^-1
B
1.33 g mol^-1
C
2.00 g mol^-1
D
2.67 g mol ^-1

Solution:

r_a = r_c/0.225 = (79.55 pm)/0.225 =353.56 pm;

$a = 2 \sqrt{2} r_{a} = 2 \sqrt{2} (353.56 p m) = 1000 p m$

There are 4 molecules of M₂X per unit cell of anti-fluorite structure. Hence

$ρ = \frac{N M}{N_{A} a^{3}} = \frac{4 (200 g {mol}^{- 1})}{(6.022 \times 10^{23} {mol}^{- 1}) {(1000 \times 10^{- 10} cm)}^{3}} = 1.33 g {mol}^{- 1}$

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