A transition metal M forms a volatile chloride which has a vapour...

$74.75 \%$ of chlorine means $74.75 \mathrm{~g}$ chlorine is present in $100 \mathrm{~g}$ of metal chloride. Weight of metal $=100 \mathrm{~g}-74.75 \mathrm{~g}$ $=25.25 \mathrm{~g}$ Equivalent weight \begin{aligned} & =\frac{\text { weight of metal }}{\text { weight of chlorine }} \times 35.5 \\ & =\frac{25.25}{74.75} \times 35.5=12 \end{aligned} Valency of metal \begin{array}{r} =\frac{2 \times \text { V.D. }}{\text { Equivalent wt. of metal }+35.5} \\ =\frac{2 \times 94.8}{12+35.5}=4 \end{array} $\therefore$ Formula of compound $=\mathrm{MCl}_4$