At $$STP,$$  a container has $$1$$ $$mole$$  of $$He,$$  $$2$$ $$mole$$  $$Ne,$$  $$3\,mole\,{O_2}$$   and $$4\,mole\,{N_2}.$$   without changing total pressure if $$2$$ $$mole$$  of $${O_2}$$  is removed. The partial pressure of $${O_2}$$  will be decreased by :

Solution :
$$\eqalign{ & {P_{{O_2}}} = \frac{3}{{10}} \times {P_T}; \cr & {\text{After removing }}2{\text{ }}mole{\text{ of }}{O_2}, \cr & P{'_{{O_2}}} = \frac{1}{{10}} \times {P_T} \cr & {\text{Decrease in partial pressure of}}\,{O_2} \cr & = \frac{{\frac{{3{P_T}}}{{10}} - \frac{{{P_T}}}{8}}}{{\frac{{3{P_T}}}{{10}}}} \times 100 \cr & = 58.33\% \cr} $$