From a certain apparatus, the diffusion rate of hydrogen has an average value of `28.7 cm^(3) s^(-1)`. The diffusion of another gas under the same condition is measured to have an average rate of `7.2 cm^(3) s^(-1)`. Identify the gas.

Options

A. Nitrogen
B. Helium
C. Argon
D. Oxygen

Answer

D

Explanation

According to Grahams' low of diffusion, `(r_(1))/(r_(2))=sqrt((rho_(2))/(rho_(1)))or (r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))`
Here, `r_(1)=` duffusion rate of hydrogen `=28.7cm^(3)s^(-1).`
`r_(2)=` diffusion rate of unknown gas `=7.2cm^(3)s^(-1).`
`M_(1)=` molecular mass of hydrogen = 2 g
`therefore(28.7)/(7.2)=sqrt((M_(2))/(2))or M_(2)=((28.7)/(7.2))^(2)xx2~~32g`
32 g is molecular mass of oxygen.