Answer:
total mass will be = = 1.207g
Explanation:
First what is given
Pressure P= 0.988 atm Room TemperatureT = 23.5°C= 296.5 K
Volume V= 1.042 L
Nitrogen in air is 80 % (moles number) = 0.8
Ideal gas constant R = 0.0821 L atmmol-1K-1
Given mass of N = 14.01 g/mol
Given mass of oxygen O = 16.00 g/mol
Total number of moles = ?
So first we have to find the total number of moles by using formula
Total number of moles
n = PV/ RT
adding the values
moles n= 0.988atm x 1.042L / (0.0821L-atm/mole-K x 296.6K)
= 1.0294 / 24.35
= 0.042 moles (total number of moles)
So by using Nitrogen percentage
Moles of nitrogen = total moles x 80/100
= 0.042moles x 0.8
So moles of O2= Total moles – moles of N2
= 0.042moles - 0.034moles
Moles of O2 = 0.008moles
Now for finding the mass of the N2 and oxygen
Mass of Nitrogen N2 = no of moles x molar mass
= 0.034 x 28
= 0.952 g
Mass of oxygen O2 = no of moles x molar mass
= 0.008 x 32
= 0.256 g
total mass will be = Mass of Nitrogen N2 + Mass of oxygen O2
=0.952 g + 0.256 g
=1.207g
