Question

A gaseous mixture contains ch4 and c2h6 in equimolar proportion

Answer 1

Suppose we take 1 mole of CH4 Weight=16gm And 1mole C2H6 weight =30gm Now, 1 mole of both CH4 and C2H6 occupy 22.4L Therefore when mixed Total weight of mixture=16+30=46 Total volume of mixture=22.4+22.4=44.8L 44.8L of mixture weigh 46g So,2.24L of mixture weighs=( 46÷44.8)×2.24=2.3g

Answer 2

Molar mass of {CH}_{4} = 16 gm

Molar mass of {C}_{2}{H}_{6} = 30 gm

1 mole of {CH}_{4} occupy = 22.4 L

1 mole of {C}_{2}{H}_{6} occupy = 22.4 L

Total molar mass = 16+30 => 46 gm

Total space occupied = 22.4 + 22.4 = 44.8 L

Using unitary method ,

44.8 L has a molecular mass = 46

:. 2.24 L has a molecular mass = 46/44.8 × 2.24

= 2.3 L

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