Question

A solution contains 0.18 g per mL of a substance X whose molecular weight is approximately 68000 . It is found that 0.27 mL of oxygen at 760 mm and $37^{\circ} C$ will combine with the amount of X contained in 1.0mL of the solution. How many molecules of oxygen will combine with one molecule of X?

Answer 1

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.