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) Molecular mass of nitrogen = 28 u
1 mole of nitrogen = 28 g = 22.4 L at STP, i.e., 28 g of nitrogen occupy 22.4 L at STP
`therefore" 14 g of nitrogen will occupy"=(22.4)/(28)xx14=11.2"L at STP"`
(ii) 1 mole of carbon dioxide have volume = 22.4 L at STP
`therefore "1.5 moles of carbon dioxide will occupy volume"=(22.4)/(1)xx1.5L=33.6"L at STP"`
(iii) 1 mole of `O_(2)` molecules = `6.022xx10^(23)" molecules"=22.4" L at STP"`
i.e., `6.022xx10^(23)` molecules of oxygen will occupy 22.4 L at STP
`10^(21)` molecules of oxygen will occupy `=(22.4)/(6.022xx10^(23))xx10^(21)"L at STP"`
`=3.72xx10^(-2)"L at STP"=3.72xx10^(-2)xx10^(3)"mL at STP"=37.2"mL at STP"`