An element X has a body centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm$-$3. The number of atoms present in 300 g of the element X is _______________.
Given : Avogadro constant, NA = 6.0 $\times$ 1023 mol$-$1.
Solution
$\rho=\frac{Z \times M}{a^{3} \times N_{\mathrm{A}}}$
<br/><br/>
$Z=2 \text { for } b c c$
<br/><br/>
$$
\begin{aligned}
& 5 \mathrm{~g} / \mathrm{cm}^{3}=\frac{2 \times M}{\left(200 \times 10^{-10} \mathrm{~cm}\right)^{3} \times 6.0 \times 10^{23}} \Rightarrow M=12 \mathrm{~g}
\end{aligned}
$$
<br/><br/>
$12 \mathrm{~g}$ of element contain $=N_{\mathrm{A}}$ atoms
<br/><br/>
$300 \mathrm{~g}$ of element contains $=N_{\mathrm{A}} \times \frac{300}{12}=25 N_{\mathrm{A}}$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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