Q20 / 160 · AP EAMCET 2024 (20 May Shift 2) · Chemistry
The density of $\beta$ - Fe is $7.6 \mathrm{~g} \mathrm{~cm}^{-3}$. It crystallizes in cubic lattice with $\mathrm{a}=290 \mathrm{pm}$. What is the value of Z ? $(\mathrm{Fe}=56$ $\left.\mathrm{g} \mathrm{mol}^{-1} ; \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)$
- Correct answerA$2$
- B$1$
- C$4$
- D$6$
Solution
$\operatorname{Density}(\rho)=\frac{Z \times M}{N_A \times a^3}$
Given, $\begin{aligned} & \rho=7.6 \mathrm{~g} / \mathrm{cm}^3 \\ & a=290 \times 10^{-12} \mathrm{~m} \\ & =290 \times 10^{-10} \mathrm{~cm} \end{aligned}$ $\begin{aligned} & \therefore \quad \frac{7.6 \times 6.023 \times 10^{23} \times\left(290 \times 10^{-10}\right)^3}{56}=Z \\ & \therefore \quad Z=2 \end{aligned}$
Given, $\begin{aligned} & \rho=7.6 \mathrm{~g} / \mathrm{cm}^3 \\ & a=290 \times 10^{-12} \mathrm{~m} \\ & =290 \times 10^{-10} \mathrm{~cm} \end{aligned}$ $\begin{aligned} & \therefore \quad \frac{7.6 \times 6.023 \times 10^{23} \times\left(290 \times 10^{-10}\right)^3}{56}=Z \\ & \therefore \quad Z=2 \end{aligned}$
Hence, the correct answer is option: 'A'.