Q72 / 180 · NEET 2013 (All India) · Chemistry
A metal has a fcc lattice. The edge length of the unit cell is $404 \mathrm{pm}$. The density of the metal is $2.72 \mathrm{~g} \mathrm{~cm}^{-3}$. The molar mass of the metal is ( $N_A$ Avogadro's constant $=6.02 \times 10^{23} \mathrm{~mol}^{-1}$ )
- A$40 \mathrm{~g} \mathrm{~mol}^{-1}$
- B$30 \mathrm{~g} \mathrm{~mol}^{-1}$
- Correct answerC$27 \mathrm{~g} \mathrm{~mol}^{-1}$
- D$20 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution
Given, cell is fcc, so $Z=4$ Edge length, $a=404 \mathrm{pm}=4.04 \times 10^{-8} \mathrm{~cm}$ Density of metal, $d=2.72 \mathrm{~g} \mathrm{~cm}^{-3}$ $N_A=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Molar mass of the metal, $M=$ ?
We know that
density, $d=\frac{Z \times M}{2^3 \cdot N_A}$
$$
\begin{aligned}
\therefore M & =\frac{d \cdot a^3 \cdot N_A}{Z} \\
& =\frac{2.72 \times\left(4.04 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}}{4} \\
& =27 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.