180 mL of a hydrocarbon diffuse through a porous membrane in 15 minutes, while 120 mL of SO_2 under identical conditions diffuse in 20 minutes. What is the molecular mass of the hydrocarbon ?

180 mL of a hydrocarbon diffuse through a porous membrane in 15 minute

1.	180 mL of a hydrocarbon diffuse through a porous membrane in 15 minutes, while 120 mL of SO_2 under identical conditions diffuse in 20 minutes. What is the molecular mass of the hydrocarbon ?
Answer» SOLUTION :Rate of diffusion of hydrocarbon `(r_(1)) = 180//15 = 12` mL/MIN Rate of diffusion of `SO_(2)(r_(2)) = 120/20 = 6 mL`/min Molecular mass of `SO_(2) (M_(2)) = 32 + (2 xx 16) = 64` According to Graham.s LAW of diffusion, `r_(1)/r_(2) = sqrt(M_(2)/M_(1))` Substituting the values, we get `12/6 = sqrt(64/M_(1))` or `M_(1) = (6 xx 6 xx 64)/(12 xx 12) = 16` Hence, the molecular mass of the hydrocarbon is 16.