0.76 g of a mixture of hydrogen and oxygen gases has a volume of 2L temperature of 300 K and pressure of `10^5 N.m^-2`. Find out the individual masses of hydrogen and oxygen in the mixture.

Explanation

Let the number of moles of hydrogen and oxygen be `n_1` and `n_2` respectively
Then the pressure of the mixture
`p=(n_1RT)/V+(n_2RT)/V=(n_1+n_2)(RT)/V`
or,`n_1+n_2=(pV)/(RT)=(10^5times(2times10^-3))/(8.3 times 300)=0.08`........(1)
Now mass of hydrogen gas `=2n_1` and mass of oxygen gas =`32 n_2`.
So, `2n_1+32n_2=0.76`
or,`n_1+16n_2=0.38`......(2)
(2)-(1) gives
`15n_2=0.3 or, n_2=0.02`
Then from (1) , `n_1=0.08-0.02=0.06`
`therefore` Mass of hydrogen= `2times 0.06=0.12 g` and mass of oxygen =`32 times 0.02 =0.64 g`