Molecular weights of two gases are 64 annd 100 respectively. If rate of diffusion of first gas be 15 `mL*sec^(-1)`, then what is the rate of diffusion of the other?

Explanation

According to Graham's law of diffusion, `(r_(A))/(r_(B))=sqrt((M_(B))/(M_(A)))`
Here, `M_(A)=64,M_(B)=100,r_(A)=15mL*s^(-1),r_(B)=?`
`therefore (15mL*s^(-1))/(r_(B))=sqrt((100)/(64))=(10)/(8)" "therefore r_(B)=12mL*s^(-1)`
`therefore`The rate of diffusion of the other gas=12 `mL*s^(-1)`.