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Explanation

(a) Let `p` : pressure of air in barometer
`pV =` constant
`p xx 50 = 75 xx 2 rArr p = 3" cm of Hg"`
mercury column descends by `3 cm`.
(b) The pressure of oxygen in the space above the mercury level `= 5" cm of Hg"`
`P = rho_(m)g h = 13.6 xx 10^(3) xx 10 xx (5)/(100) = 6800 N//m^(2)`
`V = A(25 + 5) = 2 xx 10^(-4) xx 30 xx 10^(-2) = 60 xx 10^(-6)m^(3)`
`T = 27 + 273 = 300 K`
Number of moles of oxygen
`n = (PV)/(R T) = (6800 xx 60 xx 10^(-6))/(8.3 xx 300) = 16.4 xx 10^(-5)`
Mass of oxygen
`M = nM_(0) = 16.4 xx 10^(-5) xx 32 g`
`= 5.24 xx 10^(-3) g`