The volume of $$0.0168\,mol$$   of $${O_2}$$ obtained by decomposition of $$KCl{O_3}$$  and collected by displacement of water is $$428$$  $$mL$$  at a pressure of $$754\,mm\,Hg$$   at $${25^ \circ }C.$$  The pressure of water vapour $${25^ \circ }C.$$

Solution :
$$\eqalign{ & {\text{Volume of }}\,0.0168{\text{ }}moles{\text{ at }}STP \cr & = 0.0168 \times 22400 = 376.3mL. \cr & \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}} \cr & {\text{or}}\,\frac{{760 \times 376.3}}{{273}} \cr & = \frac{{{P_2} \times 428}}{{298}} \cr & {\text{or}}\,\,{P_2} = 730\,mm \cr} $$