Calculate the volume at N.T.P occupied by
(i) 14 g of nitrogen
(ii) 1.5 gram moles of carbon dioxide
(iii) `10^(21)` molecules of oxygen.

Answer

(i) 11.2 L (ii) 33.6 L (iii) `37.2 cm^(3)`

Explanation

(i) 28.0 g of nitrogen at N.T.P occupy = 22.4 L
14.0 g of nitrogen at N.T.P occupy `= (14.0)/(28.0)xx22.4 = 11.2 L`
(ii) 1.0 mole of `CO_(2)` at N.T.P occupy = 22.4 L
1.5 moles of `CO_(2)` at N.T.P occupy at N.T.P `= 22.4 xx 1.5 = 33.6 L`
(iii) `6.022 xx 10^(23)` molecules of oxygen at oxygen at N.T.P occupy `= 22400 cm^(3)`
`10^(21)` molecules of oxygen at N.T.P occupy `= (10^(21))/(6.022xx10^(23))xx22400=37.2 cm^(3)`.