Solved Sample Paper 3 — Question 14

(a) Given,

Vapour density = 11

Molecular weight = 2 x vapour density = 2 x 11 = 22 g

Volume of 22 g of gas = 22.4 L

∴ Volume of 19 g of gas = $\dfrac{22.4}{22}$ x 19 = 19.34 L

Hence, volume occupied by 19 g of the gas at STP = 19.34 L

(b) Molecular formula of butane is C₄H₁₀

∴ Ratio of C and H is 4 : 10

Simple ratio is 2 : 5

Hence, empirical formula = C₂H₅