A diatomic molecule $$\text{X}_2$$ has a body-centred cubic (bcc) structure with a cell edge of $$300\,\text{pm}$$. The density of the molecule is $$6.17\,\text{g cm}^{-3}$$. The number of molecules present in $$200\,\text{g}$$ of $$\text{X}_2$$ is: (Avogadro constant $$(N_A) = 6 \times 10^{23}\,\text{mol}^{-1}$$)

We make use of the well-known relation that connects density $$\rho$$ of a crystalline solid with its unit-cell parameters:

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

Here $$Z$$ is the number of formula units (molecules, in the present case) per unit cell, $$M$$ is the molar mass of the formula unit, $$N_A$$ is Avogadro’s constant, and $$a$$ is the edge length of the cubic unit cell.

The question states that the solid has a body-centred cubic (bcc) structure. In a bcc lattice there is one lattice point at each of the eight corners and one at the body centre, so the total number of lattice points per cell is two. Because each lattice point is occupied by one molecule of $$\text{X}_2$$, we have

$$Z \;=\; 2$$

The edge length is given as $$300\,\text{pm}$$. First we convert this to centimetres because the density is in $$\text{g cm}^{-3}$$:

$$1\,\text{pm} = 10^{-12}\,\text{m} = 10^{-10}\,\text{cm}$$

So

$$a \;=\; 300\,\text{pm} = 300 \times 10^{-10}\,\text{cm} = 3.0 \times 10^{-8}\,\text{cm}$$

The volume of the cubic unit cell is therefore

$$a^{3} = \bigl(3.0 \times 10^{-8}\,\text{cm}\bigr)^{3} = 27 \times 10^{-24}\,\text{cm}^{3} = 2.7 \times 10^{-23}\,\text{cm}^{3}$$

The density is provided as $$\rho = 6.17\,\text{g cm}^{-3}$$, and Avogadro’s constant is $$N_A = 6 \times 10^{23}\,\text{mol}^{-1}$$. Substituting all known quantities into the density formula gives

$$6.17 \;=\; \dfrac{2\,M}{\bigl(6 \times 10^{23}\bigr)\,\bigl(2.7 \times 10^{-23}\bigr)}$$

We now isolate $$M$$ algebraically. Multiplying both sides by $$6 \times 10^{23}\,\times\,2.7 \times 10^{-23}$$ and then dividing by $$2$$ we obtain

$$M \;=\; \dfrac{6.17 \times \bigl(6 \times 10^{23}\bigr) \times \bigl(2.7 \times 10^{-23}\bigr)}{2}$$

Notice that $$10^{23} \times 10^{-23} = 10^{0} = 1$$, so the powers of ten cancel out. Hence only the numerical factors remain:

$$M \;=\; \dfrac{6.17 \times 6 \times 2.7}{2}$$

First multiply $$6 \times 2.7 = 16.2$$, then multiply by $$6.17$$:

$$6.17 \times 16.2 = 99.954$$

Finally divide by $$2$$:

$$M \;=\; \dfrac{99.954}{2} \approx 49.977 \,\text{g mol}^{-1}$$

For practical purposes we round this to

$$M \approx 50 \,\text{g mol}^{-1}$$

This is the molar mass of one $$\text{X}_2$$ molecule. We are asked for the number of molecules in $$200\,\text{g}$$ of the substance. The number of moles present is

$$n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{200\,\text{g}}{50\,\text{g mol}^{-1}} = 4\,\text{mol}$$

The number of molecules equals the number of moles multiplied by Avogadro’s constant:

$$N_{\text{molecules}} = n\,N_A = 4 \times N_A = 4\,N_A$$

Hence, the correct answer is Option C.