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Calculate the number of atoms present in unit cell of an element having molar mass $190 \mathrm{~g} \mathrm{~mol}^{-1}$ and density $20 \mathrm{~g} \mathrm{~cm}^{-3}$.
$\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=38 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
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$\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=38 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
Solution:
2081 Upvotes
Verified Answer
The correct answer is:
$4$
Density $(\rho)=\frac{M \mathrm{n}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}}$
$\begin{aligned}
& 20=\frac{190 \times \mathrm{n}}{38} \\
& \mathrm{n}=\frac{20 \times 38}{190}=4
\end{aligned}$
$\begin{aligned}
& 20=\frac{190 \times \mathrm{n}}{38} \\
& \mathrm{n}=\frac{20 \times 38}{190}=4
\end{aligned}$
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