An element with molar mass 2.7×10^–2kgmol^–1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10³kgm^–3, the radius of the element is approximately ___ × 10^–12m (to the nearest integer).

Given, Molar mass, Edge length, density
We need to find the radius of the element is approximately, from the given option.

$\begin{array}{l}\mathrm{d}=\frac{2\left(\frac{\mathrm{M}}{{\mathrm{N}}_4}\right)}{{\mathrm{a}}^3}\\ 2.7\times {10}^3=2\frac{\left(\frac{2.7\times {10}^{-4}}{6\times {10}^{21}}\right)}{{\left(405\times {10}^{-87}\right)}^1}\\ 2.7\times {10}^3=2\frac{\left(2.7\times {10}^{-7}\right)}{6\times {10}^3{\left(4.05\times {10}^{-10}\right)}^3}\\ 2.7\times {10}^3=2\frac{\left(2.7\times {10}^{-2}\right)}{6\times {10}^3\times 66.43\times {10}^{-36}}\\ 3.90=z\\ z\approx 4\text{ structure is foc }\\ \frac{a}{\sqrt{2}}=2\mathrm{r}\\ r-\frac{a}{2\sqrt{2}}-\frac{\sqrt{2}a}{4}-\frac{1.414\times 405\times {10}^{-12}}{4}\\ r=143.16\times {10}^{-12}\end{array}$

Therefore, the correct answer is 143.16 nearest integer 143.00