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At 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%` |
Answer» Correct Answer - 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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