The vapour pressure of pure benzene is 640 mm at 298 K. A solution of a solute in benzene shows a vapour pressure of 630 mm at the same temperature. Then the mole fraction of the solute is

Benzene vapor pressure ($p_o$)at 293K=640mm
Benzene solution vapor pressure at 293K= 630mm
Molality = ?
$$\begin{gathered} \frac{{\mathrm{p}}_0-\mathrm{p}}{{\mathrm{p}}_0}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}=\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}\times \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{A}}}\times \frac{1000}{1000} \\ \mathrm{Where}, \frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}\times \frac{1}{{\mathrm{W}}_{\mathrm{A}}}\times 1000=\text{ Molality } \\ \begin{array}{l}\Rightarrow \frac{{\mathrm{p}}_0-\mathrm{p}}{{\mathrm{p}}_{\circ }}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}=\text{ Molality }\times {\mathrm{M}}_{\mathrm{A}}\times \frac{1}{1000}\\ \Rightarrow \frac{640-630}{640}\times \frac{1000}{78}=\text{ Molality }\\ \Rightarrow \text{ Molality }=\frac{640-630}{640}\times \frac{1000}{78}=0.20\end{array} \\ \mathrm{Molality} = 0.20 \end{gathered}$$

Hence, the correct option is: (C) 0.20