Question

A mixture containing hydrogen and ammonia has a volume of 183.0 cm3 at 0.00 C and 1.0 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 vessel to 0.00 C and 1 atm, the volume is 80.0 cm3. Using the ideal gas law, calculate the mole fraction of NH3 in the original mixture.

Answer 1

H2 + NH3 is present

Vtotal = 183 cm3 = 183 mL = 0.183 L

T1 = 0°C = 273.1 K; P1 = 1 atm

assume all ammonia is recovered as liquid....

assume remaning gas is H2 gas

data:

P2 = 1 atm, T2 = 0°C = 273.15 K; V2 = 80 cm3 = 80 mL = 0.08 L

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

PV = nRT

n = PV/(RT)

n = (1)(0.08)/(0.082*273.15)

n = 0.0035716 moles of H2

now, from previous data

total mol initially

P1V1 = nRT1

n = P1V1/(R*T1)

n = (1)(0.183)/(0.082*273.15)

n = 0.008170

total mol = 0.008170

mol of ammonia = 0.008170 - 0.0035716 = 0.0045984

then

y-ammonia = mol of ammonia / Total mol = (0.0045984) / 0.008170 = 0.56283

then, th mol fraction of NH3 in original mix --> 0.56283