Two gases A and B having molecular weights 60 and 45 respectively are enclosed in a vessel . The weight of A is 0.50g and that of B is 0.2g. The total pressure of the mixture is 750mm. Calculate partial pressure of the two gases.
Answers
Answer:
Given
Weight of gas A = 0.50 g
Molecular weight of gas A = 60
Weight of gas B = 0.2 g
Molecular weight of gas B = 45
Pm = 750mm
From Daltons's law of partial pressure
P′A=Pm×PA′=Pm× mole fraction of A
=750×0.560(0.560)+(0.245)=750×0.560(0.560)+(0.245)
=489.23mm=489.23mm
Now Pm=P′A+P′BPm=PA′+PB′
P′B=Pm−P′APB′=Pm−PA′
=750 - 489.23 = 260.77 mm
No. of moles of dioxygen = 70.6 / 32 = 2.21 mol
No. of moles of neon = 167.5 / 20 = 8.375 mol
Mole fraction of dioxygen = 2.21 / (2.21 + 8.375)
= 0.21
Mole fraction of neon = 8.375 / (2.21 + 8.375)
= 0.79
Alternatively,
Mole fraction of neon = 1 - 0.21 = 0.79
Partial pressure of gas = Mole fraction×total pressure
=>partial pressure of oxygen= 0.21 x 25 bar= 5.25bar
partial pressure of neon = 19.75 bar
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