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JEE MainChemistryToughPhysical Chemistry

Ferrous oxide has a cubic structure and each edge of the unit
cell is 5.0 Assuming density of the oxide as 4.0 g - cm^-3, then
the number of Fe^2+ and O^2- ions present in each unit cell will be

Show answer and solution
Let the units of ferrous oxide in a unit cell = n, molecular
weight of ferrous oxide
(FeO)=56+16=72gmol−1\left( F_{e}O \right) = 56 + 16 = 72gmol^{- 1}
 weight of n units =72×n6.023×1023= \frac{72 \times n}{6.023 \times 10^{23}}
 Volume of one unit = (length of corner )^3
=(5A)3=125×10−24cm3= (5A)^{3} = 125 \times 10^{- 24}{cm}^{3}
 Density = wt. of cell  volume \ Density\ = \frac{\ wt.\ of\ cell\ }{\ volume\ }
4.09=72×n6.023×1023×125×10−244.09 = \frac{72 \times n}{6.023 \times 10^{23} \times 125 \times 10^{- 24}}
n=3079.2×10−172=42.7×10−1=4.27≈4n = \frac{3079.2 \times 10^{- 1}}{72} = 42.7 \times 10^{- 1} = 4.27 \approx 4

Source: Grass-G / CH-12 Chemistry Paper 2 Nov..docx

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