At what temperature, the rate of effusion of `N_(2)` would be 1.625 times that of `SO_(2)` at `50^(@)C` ?

Options

A. 110 K
B. 173 K
C. 373 K
D. 273 K

Answer

C

Explanation

`(r_(N_(2)))/(r_(SO_(2)))=(V_("rms")N_(2))/(V_("rms")SO_(2))=sqrt((T_(N_(2)))/(T_(SO_(2))).(M_(SO_(2)))/(M_(N_(2))))=sqrt((T_(N_(2)))/(323)xx(64)/(28))`
`1.625=sqrt((T_(N_(2)))/(323).(16)/(7))`
`T_(N_(2))=((1.625)^(2)xx323xx7)/(16)=373K`