We have 22.4 L at STP gas mixture of He and N2.
The weight of this gas mixture is 8.8 g
What is partial pressure of He (in Pa) ?
I found the answer but not sure
So :
22.4 L gas at STP will contains 1 mole of the gas
Mole He + mole H2 = 1 mole
A + B = 1 mole
Mass He + mass N2 = 8.8 g
A .MrHe + B .MrN2 = 8.8 g
2 g/mole A + 28 g/mole B = 8.8 g
2A + 28B = 8.8 moles
26 B = 6.8 moles
B = 68/260 = 34/130 moles
A = 96/130 moles
At STP , pressure is 1 atm, means the gas mixture has 1 atm pressure or 101,325 Pa.
Partial pressure of He =
( mole He/total mole ) × total pressure
= ( 96/130 ) × 101,325 Pa
= 74,824 Pa
Is it true?
The answer 80,000 Pa though