Correct Answer - A
`6.02 xx10^(22)` molecules of each `N_(2),O_(2)` and `H_(2)=(6.02xx10^(22))/(6.02 xx 10^(23))` moles of each
Weight of mixture `=` weight of 0.1 mole `N_(2) + `weight of 0.1 mole `H_(2)+` weight of 0.1 mole of `O_(2)`
`(28 xx 0.1) +(2xx0.1)+(32 xx0.1)=6.2 ` gm