Assume that air is 21%${\mathrm{O}}_2$ and 79% ${\mathrm{N}}_2$  by volume. If the barometric pressure is 740 mm, the partial pressure of ${\mathrm{O}}_2$ is closest to which one of the following

No.of moles of ${\mathrm{O}}_2 =21\% = 0.21$

${\mathrm{N}}_2=79\%=0.79$

Total moles of air mixture = 0.21 + 0.79 =1.00

Mole fraction of oxygen to be calculated as follows:

${\mathrm{X}}_{{\mathrm{O}}_2}=\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{O}}_2}{\mathrm{Total} \mathrm{moles} \mathrm{of} \mathrm{mixture}}$

$\frac{0.21}{1.00}=0.21 tc " total no. of moles = 1 "tc " "$Then the partial pressure of oxygen is

${\mathrm{P}}_{{\mathrm{O}}_2}={\mathrm{P}}_{\mathrm{total} }\times {\mathrm{X}}_{{\mathrm{O}}_2}$

     = 740 mm   × 0.21

    = 155 mm og Hg = 740 × 0.21=155.4 mm of Hg tc " = 740 × 0.21 =155.4."