The Graham's law states that ''at constant pressure and temperature the rate of diffusion or effusion of a gas is inversely proportional to the squar root of its density Rate of diffusion `prop (1)/(sqrt(d))`
If `r_(1)` and `r_(2)` represent the rates of diffusion of two gases and `d_(1)` and `d_(2)` are their respective densities, then
`r_(1)/(r_(2))=sqrt((d_(2))/(d_(1)))`
`r_(1)/(r_(2)) =sqrt((M_(2))/(M_(1))) xx P_(1)/(P_(2))`
`(V_(1)xxt_(2))/(V_(2)xxt_(1)) = sqrt((d_(2))/(d_(1))) = sqrt((M_(2))/(M_(1)))`
`V prop n` (where n is no of moles)
`V_(1) prop n_(1)` and `V_(2) prop n_(2)`
If some moles of `O_(2)` diffuse in 18 sec and same moles of other gas diffuse in `45sec` then what is the molecular weight of the unknown gas ? .

Options

A. `(45^(2))/(18^(2))xx32`
B. `(18^(2))/(45^(2))xx32`
C. `(18^(2))/(45^(2)xx32)`
D. `(45^(2))/(18^(2)xx32)`

Answer

B

Explanation

`r_(O_(2))/r_(X)=sqrt(M_(X)/M_(O_(2))),` As volume are in the same ratio as the number of moles
`(n//18)/(n//48)=sqrt(M_(X)/(32)),(45)/(18)=sqrt(M_(X)/(32))`
`M_(X)=(45^(2))/(18^(2))xx32`