I don't know but I'm wasting 5 seconds of your time you can't take back sorry
Explanation:
require an emergency support immediately
Answer:
Mole fraction of nitrogen = 0.52
Explanation:
Given data:
Temperature = 31.2 °C
Pressure = 870.2 mmHg
Volume = 15.1 L
Mass of mixture = 24.1 g
Mole fraction of nitrogen = ?
Solution:
Pressure conversion:
870.2 /760 = 1.12 atm
Temperature conversion:
31.2 + 273 = 304.2 K
Total number of moles:
PV = nRT
n = PV/RT
n = 1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K
n = 16.9 L.atm. /25 L.atm. mol⁻¹
n = 0.676 mol
Number of moles of nitrogen are = x
Then the number of moles of CO₂ = 0.676 - x
Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)
24.1 = 28x + ( 29.7 -44x)
24.1 - 29.7 = 28x - 44x
-5.6 = -16 x
x = 0.35
Mole fraction of nitrogen:
Mole fraction of nitrogen = moles of nitrogen / total number of moles
Mole fraction of nitrogen = 0.35 mol / 0.676 mol
Mole fraction of nitrogen = 0.52
I think that this sentence is false
When P1/P2 = C1/C2
and C is the molarity which = moles/volume
so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]
P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L
so, Mw and 1.5 L will cancel out:
∴P1/P2 = mass1 / mass2
∴ mass 2 = mass1*(P2 / P1)
= 0.278g * (78 bar / 62 bar)
= 0.35 g
∴ the quantity of argon that will dissolve at 78 bar = 0.35 g