The density of chromium is 7.2 gm cm^–3 and crystallises in a body centred cubic lattice with edge length 290pm. The number of atoms present in 30.0 gm of the sample is

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{a}}} \\ \mathrm{Z}= \mathrm{no}. \mathrm{of} \mathrm{atoms} \mathrm{per} \mathrm{unit} \mathrm{cell}=2 \\ \mathrm{M}=\mathrm{Molar} \mathrm{mass} \\ \mathrm{a}=\mathrm{edge} \mathrm{length}=290\mathrm{pm}=290\times {10}^{-10}\mathrm{cm} \\ 7.2=\frac{2\times \mathrm{M}}{{\left(290\times {10}^{-10}\right)}^3\times \left(6.023\times {10}^{23}\right)} \\ \mathrm{M}=52.68\mathrm{g} \\ \mathrm{no}. \mathrm{of} \mathrm{moles}=\frac{30}{52.68}=0.569\mathrm{moles} \\ \mathrm{no}. \mathrm{of} \mathrm{atoms}=0.569\times 6.023\times {10}^{23}=3.4\times {10}^{23} \end{gathered}$$

Therefore, the correct option is C