1.

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

Answer»

135K
373K
546 K
303K

Solution :`( r ( N_(2)))/( r(SO_(2))) = 1.625 =SQRT(((3RT( N_(2)))/(28))/((3Rxx 323)/(64)))`
`= 1.625 = sqrt(( 16 xx T ( N_(2)))/( 7 xx 323))`
`:. T ( N_(2)) = 373 K`


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