A mixture of `NH_(3)` (g) and `N_(2) H_(4)` (g) is placed in a sealed container at 300 K. The pressure within the container is 0.6 atm. The container is heated to 1000 K where the gases undergo complete decomposition as:
`2NH_(3) (g) to N_(2) (g) + 3H_(2) (g)` and `N_(2) H_(4) (g) to N_(2) (g) + 2 H_(2) (g)`
The pressure of the container at this stage becomes 4.8 atm. The mole per cent of `NH_(3) (g)` in the original mixture was

Options

A. 40
B. 50
C. 60
D. 70

Answer

C

Explanation

Let x be the mole fraction of `NH_(3)` in the original mixture .
Amount of gas to start with = 1 mol
Amount of gas at the end = 2x mol + (1- x) mol = (3-x) mol Since pV = nRT , we have `(n_(1) T_(1))/(n_(2) T_(2)) = (p_(1))/(p_(2)) implies ((1)/(3 - x)) ((300)/(1000)) = (0.6)/(4.8) = (1)/(8)`