Step I. Calculation of no. of mole of oxygen.
According to gas equation, `PV = nRT`
`P = 1 xx 10^(-12) "bar", V = 1 dm^(3), T = 273 + 0 = 273 K`
`n = (PV)/(RT) = ((1 xx 10^(-12) "bar") xx (1 dm^(3)))/((0.83 dm^(3) "bar" K^(-1) mol^(-1)) xx (273 K)) = 4.41 xx 10^(-14) mol`
Step II. Calculation of no. of molecular of oxygen
`1` mole of oxygen contain of `= 6.022 xx 10^(23)` molecules
`4.41 xx 10^(-14)` mole of oxygen contain `= 6.22 xx 10^(23) xx 4.41 xx 10^(-14) = 2.66 xx 10^(10)` molecules.