1.

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

Answer»

110 K
173 K
373 K
273 K

Solution :`(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`


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