An element with molar mass $$2.7 \times 10^{-2}$$ kg mol$$^{-1}$$ forms a cubic unit cell with edge length 405 pm. If its density is $$2.7 \times 10^3$$ kg m$$^{-3}$$, the radius of the element is approximately ......... $$\times 10^{-12}$$ m (to the nearest integer)

We are given the data - molar mass $$M = 2.7 \times 10^{-2}\; \text{kg mol}^{-1}$$, edge length of the cubic unit cell $$a = 405\ \text{pm} = 405 \times 10^{-12}\ \text{m}$$ and density $$\rho = 2.7 \times 10^{3}\; \text{kg m}^{-3}$$. To identify the type of cubic unit cell we first connect these quantities through the density relation for crystals.

The general density formula is

$$\rho \;=\; \dfrac{Z\,M}{N_A\,a^{3}}$$

where $$Z$$ is the number of atoms per unit cell and $$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$$ is Avogadro’s constant.

Re-arranging for $$Z$$ we get

$$Z \;=\; \dfrac{\rho\,N_A\,a^{3}}{M}$$

Now we substitute the given values one by one. First we evaluate $$a^{3}$$:

$$a = 405 \times 10^{-12}\ \text{m}$$

$$a^{3} = (405 \times 10^{-12})^{3} \text{ m}^{3} = 405^{3}\times 10^{-36}\ \text{m}^{3}$$

We calculate $$405^{3}$$ explicitly:

$$405^{3} = 405 \times 405 \times 405 = 164{,}025 \times 405 = 66{,}836{,}125$$

Hence

$$a^{3} = 66{,}836{,}125 \times 10^{-36}\ \text{m}^{3} = 6.6836125 \times 10^{-29}\ \text{m}^{3}$$

Next we assemble $$\rho N_A a^{3}$$:

$$\rho\,N_A\,a^{3} = (2.7 \times 10^{3})\,(6.022 \times 10^{23})\,(6.6836125 \times 10^{-29})$$

Multiplying the numerical parts in steps,

$$2.7 \times 6.022 = 16.2594$$

$$16.2594 \times 6.6836125 \approx 108.6707$$

For the powers of ten we have $$10^{3}\times10^{23}\times10^{-29}=10^{-3}$$. So

$$\rho\,N_A\,a^{3} \approx 108.6707 \times 10^{-3} = 0.1086707$$

Now we divide by the molar mass:

$$Z = \dfrac{0.1086707}{M}\,, \qquad M = 2.7 \times 10^{-2} = 0.027$$

$$Z = \dfrac{0.1086707}{0.027} \approx 4.024 \approx 4$$

Obtaining the nearest integer value $$Z = 4$$ tells us that the unit cell is face-centred cubic (fcc).

For an fcc lattice the relation between the edge length $$a$$ and the atomic radius $$r$$ is obtained from the face diagonal. On the face diagonal we have four radii, hence

$$4r = \sqrt{2}\,a \quad\Longrightarrow\quad r = \dfrac{\sqrt{2}}{4}\,a = \dfrac{a}{2\sqrt{2}}$$

Substituting $$a = 405 \times 10^{-12}\ \text{m}$$ gives

$$r = \dfrac{405 \times 10^{-12}}{2\sqrt{2}} = \dfrac{405}{2 \times 1.414}\times 10^{-12}\ \text{m}$$

Calculating the numerator-denominator ratio,

$$\dfrac{405}{2.828} \approx 143.2$$

Hence

$$r \approx 143.2 \times 10^{-12}\ \text{m}$$

and to the nearest integer the radius is $$143 \times 10^{-12}\ \text{m}$$.

Hence, the correct answer is Option 143.