A transition metal $$M$$ forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be

The volatile chloride is in the gaseous state, so its molar mass can be obtained from its vapour density (V.D.).

For any gas, $$\text{Molar mass} = 2 \times \text{V.D.}$$

Given $$\text{V.D.} = 94.8$$, therefore
$$\text{Molar mass of the chloride} = 2 \times 94.8 = 189.6\ \text{g mol}^{-1}$$

The chloride contains $$74.75\%$$ of chlorine by mass, so the mass of chlorine present in one mole of the compound is
$$\text{Mass of Cl} = 0.7475 \times 189.6 = 141.726\ \text{g}$$

The atomic mass of chlorine is $$35.5\ \text{g mol}^{-1}$$. Hence the number of chlorine atoms in one molecule is
$$n = \frac{141.726}{35.5} \approx 3.99 \approx 4$$

Therefore the empirical (and molecular) formula of the chloride is $$MCl_4$$.

The remaining mass belongs to the metal: $$189.6 - 141.726 \approx 47.9\ \text{g mol}^{-1}$$, matching titanium, which indeed forms the volatile compound $$TiCl_4$$. This further supports the formula.

Option C which is: $$MCl_4$$