SOLUTION : Calculation of no. of mole of $X$ present in $1.0 mL$
Mass of $1 \cdot 0 mL$ of $X=$ volume $\times$ density $=1 \cdot 0 \times 0 \cdot 18 g =0 \cdot 18 g$
no. of mole $=\frac{\text { mass }}{\text { mol. wt. }}=\frac{0 \cdot 18}{68000}$ mole.
Hence no. of molecules of $X$ in $1.0 mL$ solution $=\frac{0 \cdot 18 \times N_{A}}{68000}$ molecules.
Calculation of no. of mole of oxygen
Volume of oxygen at S.T.P. $=\frac{760 \times 0.27}{(273+37)} \times \frac{273}{760} mL$
$=\frac{0 \cdot 27 \times 273}{310} mL =0.237 mL \text { at S.T.P. }$
Since $22400 mL$ at S.T.P. $=1$ mole
$\therefore \quad 0.237 mL$ at S.T.P. $=\frac{0 \cdot 237}{22400} \quad$ mole $=\frac{0 \cdot 237 \times N_{A}}{22400}$ molecules
From problem
$\frac{0 \cdot 18 \times N_{A}}{68000}$ molecules of $X$ combine with $\frac{0 \cdot 237 \times N_{A}}{22400}$ molecules of oxygen
One molecule of $X$ combines with $\frac{0 \cdot 237 \times N_{A}}{22400} \times \frac{68000}{0 \cdot 18 \times N_{A}} \approx 4$ molecules.