75 ml of gas A effuses through a pin hole in 73 seconds the same volume of `SO_(2)` under indential conditions effuses in seconds. Calculate the molecular mass of A.

Explanation

Given : Time taken for effusion of gas A=73 seconds
ime taken for effusion of `SO_(2)` gas =75 seconds
Volume of gases `(A & SO_(2))` taken =75 ml each.
Molecular mass of `SO_(2)`=
Sol. `("Effusion rate of" SO_(2))/("Effusion rate of A") = sqrt((M_(A))/(M_(SO_(2))))`
`(75//75)/(75//73)=sqrt((M_(A))/(M_(SO_(2)))), (1)/(1.027)=sqrt((M_(A))/(64))`
`:.` Molecular mass of A is `60.67`