1.

Same mass of CH_(4) and H_(2)is taken in container. The partial pressure caused by H_(2) is

Answer»

`8 //9`
`1//9`
`1//2`
1

Solution :`N_(CH_(4))=` NUMBER of MOLES of `CH_(4)=(m)/(16)`
`N_(H_(2))= ` number of moles of `H_(2) = (m)/(2)`
fractionpartial pressure of `H_(2)` is
`H_(2)=(n_(H_(2)))/(n_(H_(2))+n_(CH_(4)))=((m)/(2))/((m)/(2)+(m)/(16))=((m)/(2))/((9M)/(16))=(8)/(9)`


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