Chemistry - Solid State Question with Solution | TestHub

(B) $a=500\mathrm{pm}=5\times 10^{-8}\mathrm{cm}$ $\therefore {\mathrm{a}}^3={\left(5\times 10^{-8}\mathrm{cm}\right)}^3=125\times 10^{-24}{\mathrm{cm}}^3$ $\rho =7.5\mathrm{g}{\mathrm{cm}}^{-3},\mathrm{m}=300\mathrm{g},\mathrm{n}=2($ for bcc cell $)$ $\rho =\frac{\mathrm{nM}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}\therefore \mathrm{M}=\frac{\rho \times {\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}{\mathrm{n}}$ $\therefore M=\frac{7.5\times 125\times 10^{-24}\times 6.022\times 10^{23}}{2}=282.3\mathrm{g}{\mathrm{mol}}^{-1}$ $282.3\mathrm{g}$ of metal contains $6.022\times 10^{23}$ atoms $\therefore 300\mathrm{g}$ of metal contains $\frac{6.022\times 10^{23}\times 300}{282.3}=6.399\times 10^{23}$ atoms $\approx 6.4\times 10^{23}$ atoms