Question

What is the molecular formula of the alkane, the 5.6 litre of which weight 11 g at STP

• A. ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$
• B. ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$
• C. ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$
• D. ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

Answer 1

The correct option is C ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$

22.4L of gas = 1mol
1L= 1/22.4 mol
5.6L = 1/4 mol
let molar mass of gas is M then
5.6L = M/4gm =11gm  (given)
 so M=11*4=44gm
Hence option (C) is correct.