What volume at S.T.P is occupied by (i) 3.50 g of nitrogen? (ii) 6.022 xx 10^(21)molecules of ammonia? (iii) 0.350 moles of oxygen? (iv) 39.9 g of argon?

What volume at S.T.P is occupied by (i) 3.50 g of nitrogen? (ii) 6.0

1.	What volume at S.T.P is occupied by (i) 3.50 g of nitrogen? (ii) 6.022 xx 10^(21)molecules of ammonia? (iii) 0.350 moles of oxygen? (iv) 39.9 g of argon?
Answer» Solution :(i) The NUMBER of moles in 3.50 g of nitrogen `=(3.50)/(28.02) = 0.125` (since, the gram molecular mass of `N_(2)` is 28.02 g) `therefore` One mole of a gas occupies 22.4 L at S.T.R `therefore` Thevolumeoccupiedby0.125moles =` 0.125 xx 22.4 = 2.80 L` Hence, 3.50 g of N2 occupy a VOLUME of 2.80 litres at S.T.P Ans. (ii) One mole of a gas has 6.022 x 10 MOLECULES and occupies a volume of 22.4 L at S.T.P. `therefore` The volume occupied by `6.022 xx 10^21` molecules. (III) The volume occupied by 0.350 moles of `O_(2)` at S.T.P. = `0.350 xx 22.4 = 7.84` L (because the volume occupied by one mole at S.T.P. is 22.4 L) Hence, 0.350 moles of `O_(2)` occupy 7.84 litres at S.T.P. Ans. (iv) Argon is monoatomic in nature. `therefore` 1 gram mole of Ar = 1 g atom of Ar = 39.95 g Hence, 39.95 g (1 mole) of argon will occupy a volume of 22.4 L at S.T.P.