1.

If some moles of O_(2) diffuse in 18 sec and same moles of other gas diffuse in 45 sec then what is the molecular weight of the unknown gas

Answer»

`(45^(2))/(18^(2))xx32`
`(18^(2))/(45^(2))xx32`
`(18^(2))/(45^(2)xx32)`
`(45^(2))/(18^(2)xx32)`

Solution :`r_(O_(2)): (X)/(18) `mole `//` sec`impliesr_(g) =(x)/(45) mol//` sec
`M_(g)=M_(O_(2))((r_(O_(2)))/(r_(g)))^(2) = 32 ((x)/(18) xx(45)/(x))^(2) = 32 xx (45^(2))/(18^(2))`


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