The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp. and press.

(i) Which sample contains the maximum no. of molecules. If the temp, and pressure of gas A are kept constant, what will happen to the volume of A when the no. of molecules is doubled.

(ii) If the volume of 'A' is 5.6 dm³ at s.t.p., calculate the no. of molecules in the actual vol. of 'D' at s.t.p. [Avog no. is 6 × 10²³).

Using your answer, state the mass of 'D' if the gas is N₂O

[N = 14, O = 16]

Volume is directly proportional to the number of molecules, hence gas D will have maximum no. of molecules as its volume is maximum.

If number of molecules of gas A is doubled, the volume will also be doubled.

(ii) 1 mole contains 6 x 10²³ number of molecules and occupies 22.4 lit. vol.

Given, volume of 'A' is 5.6 dm³ at s.t.p.

∴ vol. of D will be 4 × 5.6 = 22.4 lit.
No. of molecules in 22.4 lit. of D = 6 x 10²³ (Avogadro no.)

As D is 1 mole hence, mass of 1 mole of D (N₂O) = 2(14) + 16 = 28 + 16 = 44 g

Hence, mass of N₂O = 44 g