At a certain temperature the time required for the complete diffusion of 200 mL of `H_(2)` gas is 30 minute. The time required for the complete diffusion of 50 mL of `O_(2)` gas at the same temperature will be-

Options

A. 60 mins
B. 30 mins
C. 45 mins
D. 15 mins

Answer

B

Explanation

According to Graham's law, `(V_(H_(2))//t_(1))/(V_(O_(2))//t_(2))=sqrt(M_(O_(2))//M_(H_(2)))`
or, `(V_(H_(2)))/(V_(O_(2)))xx(t_(2))/(t_(1))=sqrt((M_(O_(2)))/(M_(H_(2)))) or, (200)/(50)xx(t_(2))/(30"min")=sqrt((32)/92))=4`
`therefore t_(2)=30` min.