A container of 1 L capacity contains a mixture of 4 g of `O_(2)` and 2 g `H_(2)` at `0 .^(@)C` . What will be the total pressure of the mixture ? (a) 50 . 42 atm (b) 25 . 21 atm (C) 15 . 2 atm (d) 12 . 5 atm

Explanation

Number of moles `=(Mass)/(Molar mass)`
Moles of `N_(2)(n_(N_(2)))=(2.8)/(28)=0.1mol`
Moles of `O_(2)(n_(O_(2)))=(3.2)/(32)=0.1 mol`
Moles of `H_(2)(n_(H_(2)))=(0.5)/(2)=0.25 mol`
`:. ` Total number of mole `=n_(N_(2))+n_(O_(2))+n_(H_(2))`
`=0.1+0.1+0.25=0.45`
Partial pressure a gas is
`(Number of moles)/(Total number of moles)xxTotal pressure`
`P_(N_(2))=(0.1)/(0.45)xx4.5atm=1.0 atm`
`P_(O_(2))=(0.1)/(0.45)xx4.5 atm=1.0 atm`
`P_(H_(2))=(0.25)/(0.45)xx4.5 atm=2.5 atm`