At S.T.P 5.6 lit of acetylene gas is passed through excess of ammonical AgNO₃ solution. Then the weight of precipitate obtained is

1 mole of gas has volume 22.4 L then no of moles in 5.6 L gas will be   5.6/ 22.4
=0.25 moles
If 0.25 moles of gas weighs 60 g

then 1 mole of gas weighs= 60/0.25
=240g

Molar mass=240g

Molar mass=2×Vapourdensity
240=2×V.D
⇒V.D=120