An element M crystallises in a body centred cubic unit cell with a cell edge of $300 \,\mathrm{pm}$. The density of the element is $6.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of atoms present in $180 \mathrm{~g}$ of the element is ____________ $\times 10^{23}$. (Nearest integer)
Answer (integer)
22
Solution
$M$ is body certred cubic, $\therefore Z=2$<br/><br/>
Let mass of 1 atom of $M$ is $A$<br/><br/>
Edge length $=300 \,\mathrm{pm}$<br/><br/>
Density $=6 \mathrm{~g} / \mathrm{cm}^3$<br/><br/>
$$
\therefore 6 \mathrm{~g} / \mathrm{cm}^3=\frac{\mathrm{Z} \times \mathrm{A}}{\left(300 \times 10^{-10}\right)^3}=\frac{2 \times \mathrm{A}}{27 \times 10^{-24}}
$$<br/><br/>
$\mathrm{A}=81 \times 10^{-24} \mathrm{~g}$<br/><br/>
$\therefore$ Atomic mass $=48.6 \mathrm{~g}$<br/><br/>
$\therefore$ Mole in $180 \mathrm{~g}=\frac{180}{48.6}=3.7$ moles<br/><br/>
Atoms of $\mathrm{M}=3.7 \times 6 \times 10^{23}$
$=22.22 \times 10^{23}$ atoms
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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