A cylinder contains 68 g of Ammonia gas at STP

(i) What is the volume occupied by this gas?

(ii) How many moles of ammonia are present in the cylinder?

(iii) How many molecules of ammonia are present in the cylinder? [N-14, H-1]

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(i) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.

17 g. occupies 22.4 lit. of vol.

∴ 68 g will occupy = $\dfrac{22.4}{17}$ x 68 = 89.6 lit.

Hence, volume occupied by this gas = 89.6 lit.

(ii) 17 g = 1 mole

∴ 68 g = $\dfrac{1}{17}$ x 68 = 4 moles.

1 mole = 6 × 10²³

∴ 4 moles = 4 x 6.023 × 10²³ molecules.

Hence, Moles = 4

(iii) molecules = 4 x 6.023 × 10²³ = 2.4 x 10²⁴ molecules