(i) Gram molecular mass of methane = V.D. × 2
= 8 × 2 = 16 g
16 g of methane occupies 22.4 lit.
∴ 40 g of methane will occupy = 22.4\over16 × 40 = 56 lits.
(ii) Gram molecular mass of NaOH = 23 + 16 + 1 = 40 g
40 g of NaOH = 1 mole
∴ 160 g of NaOH = 160\over40 = 4 moles.