Question

A mixture of 0.200 g of H2, 0.800 g of N2, and 0.600 g of Ar is stored in a closed container at 2.0 atm and 10°C. What is the partial pressure of N2 in the container?

ans: 0.40 atm

please explain

Answer 1

molar mass of H2 = 2 g/mol
molar mass of N2 = 28 g/mol
molar mass of Ar = 40 g/mol
number of mole = (given mass)/(molar mass)
number of mole of H2 = 0.200/2
= 0.1 mole
number of mole of N2 = 0.8/28
= 0.029 mole
number of mole of Ar = 0.6/40
= 0.015 mole

mole fraction of N2 = (number of mole oof N2)/(total number of mole)
= 0.029/(0.1+0.029+0.015)
= 0.029/0.144
= 0.2

partial pressure = (mole fraction)*(total pressure)
partial pressure of N2 = 0.2*2
= 0.4 atm

Answer : 0.4 atm