`12 cm^3` of `SO_2 (g)` diffused through a porous membrane in 1 minute. Under similar conditions `120 cm^3` of another gas diffused in 5 minutes. The molar mass of the gas in `g"mol"^(-1)` is

Options

A. 32
B. 18
C. 44
D. 16

Answer

D

Explanation

According to law of diffusion rate(r ) `alpha (1)/(sqrtM)` (molar mass)
and (rate) `alpha ("volume (V)")/("time (t)")`
i.e., `(r_((SO_2)))/(r_((z))) = ((V // t)_((so_2)) )/((V//t)_((z)) ) = ([1 //sqrtM]_((SO_2) ))/((1//sqrt(M)_C)`
or , ` (V_((SO_2)) xx t_((z)))/(V_((z)) xx t_((SO_2))) = sqrt( (M_((z)))/(M_((SO_2))))`
Given ,
`V_((SO_2)) = 12 cm^3`
`t_((SO_2)) ` = 1 minute = ` 1 xx 60 = 60 `sec
`V_((z)) = 120 cm^3`
`t_((z)) ` = 5 minutes ` = 5 xx 60 = 300` sec
and molar mass (M) of `SO_2 = (32+ (2 xx 16) = 64`
Thus , `(12 xx300)/(120 xx 60) = sqrt((M_z)/(64) ) " or " 1/4 = (M_z)/(4)`
`therefore M_z = 64/4 = 16`
Hence , molar mass of given gas is ` = 16 g "mol"^(-1)`