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Assuming complete decomposition of `NH_(3)` and `N(2)H_(4)`
`P=0.3 atm, P=2.7 atm`
`T=300K, T=200K`
`VL, VL`
mole `%` of `NH_(3)` in original mixture is (assume both concentration same volume)

A

`25%`

B

`20%`

C

`75%`

D

`37.5%`

Text Solution

Verified by Experts

The correct Answer is:
C
Using equation
`PV=nRT`
`n_(1)`= moles of `NH_(3)`
`n_(2)`= moles of `N_(2)H_(4)`
`0.3xxV=(n_(1)+n_(2))xxRxx300`
`2.7xxV=(2n_(1)+3n_(2))xxRxx1200`
`(2n_(1)+3n_(2))/(n_(1)+n_(2))=(9/4)`
`n_(1)/n_(2)=(1/3)`
`n_(1)/n_(2)=75%`
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