What volume is occupied at N.T.P. by
(i) 1.4 g of nitrogen
(ii) `6.023 xx 10^21` molecules of oxygen
(iii) 0.2 mole of ammonia ?

Answer

A, B, D

Explanation

(i) 28 g of nitrogen at N.T.P. occupy volume = 22.4 L
1.4 g of nitrogen at N.T.P. will occupy volume
`=(22.4)/28 xx 1.4 = 1.12` L
(ii) `6.022 xx 10^(23)` molecules of `O_(2)` at N.T.P. occupy volume = 22.4 L
volume `=(22.4 xx 6.022 xx 10^(23))/(6.022 xx 10^(23))`
=0.224 L
(iii) 1 mol of `NH_(3)` occupy volume = 22.4 L
0.2 mol of `NH_(3)` occupy volume `=22.4 xx 0.2 = 4.48` L