Question

A mixture of H2 and NH3 has a volume of 137 cm^3 at 0.00°C and 1.00 atm. The mixture is cooled to the temperature of liquid nitrogen at which ammonia freezes out and the remaining gas is removed from the vessel. Upon warming the removed gas to 0.00°C and 1.00 atm, the volume is 76.0 cm^3.
1. Using ideal gas law calculate the mole fraction of NH3 in the original mixture.

Answer 1

Answer:

Explanation:

volume of H2 &NH3=137.0cm^3

V=0.1370L

T=273.1K(T=0°C+273.1K)

P=1.00atm

л(total)=л(NH3)+л(NH2)=PV/RT

When NH3 freezes ,H2 is removed &then warmed to 0.00°C then V=0.0760 T=273.1K & P=1.00 atm

лNH3=PV/RT

x(NH3)=л(NH3)/л(total)