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

1. A

0.67 g mol-1

2. B

1.33 g mol-1

3. C

2.00 g mol-1

4. D

2.67 g mol -1

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Solution:

ra = rc/0.225 = (79.55 pm)/0.225 =353.56 pm;

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

$\rho =\frac{NM}{{N}_{\mathrm{A}}{a}^{3}}=\frac{4\left(200\mathrm{g}{\mathrm{mol}}^{-1}\right)}{\left(6.022×{10}^{23}{\mathrm{mol}}^{-1}\right){\left(1000×{10}^{-10}\mathrm{cm}\right)}^{3}}=1.33\mathrm{g}{\mathrm{mol}}^{-1}$

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