Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the $\mathrm{FeO}$ in the crystal is $4.0 \mathrm{~g} \mathrm{~cm}^{-3}$, then the number of $\mathrm{FeO}$ units present per unit cell is __________. (Nearest integer)
Given: Molar mass of $\mathrm{Fe}$ and $\mathrm{O}$ is 56 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively. $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$
Answer (integer)
4
Solution
<p>$\mathrm{d=\frac{z\times m}{a^3}}$</p>
<p>$\mathrm{4=\frac{z\times72}{6\times10^{23}(5\times10^{-8})^3}}$</p>
<p>$\mathrm{4=\frac{z\times72}{6\times125\times10^{-1}}}$</p>
<p>$=\mathrm{z \approx4}$</p>
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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