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.

Explanation

(i) Volume of 14 g of nitrogen at N.T.P
Molar mass of nitrogen `(N_(2))=28` g
28 g of `N_(2)` at N.T.P. occupy volume `= 22.4 L`
14 g of `N_(2)` at N.T.P occupy volume `= ((22.4L))/((28g))xx(14g)=11.2L`
(ii) Volume of 1.5 gram moles of carbon dioxide at N.T.P
1.0 gram mole of `CO_(2)` at N.T.P occupy volume `=22.4L`
1.5 gram moles of `CO_(2)` at N.T.P occupy volume `=22.4 xx 1.5 = 33.6`L
(iii) Volume of `10^(21)` molecules of oxygen at S.T.P
We know that, 22.4 L is the volume of `6.022 xx 10^(23)` molecules of a gas at N.T.P
Thus, `6.022 xx 10^(23)` molecules of `O_(2)` at N.T.P occupy volume = 22.4 L
`10^(21)` molecules of `O_(2)` at N.T.P occupy volume `= ((22.4L)xx10^(21))/(6.022xx10^(23))=3.72xx10^(-2)L`
`=3.72xx10^(-2)xx10^(3)mL=37.2mL`.