Question

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In a mixture of argon and hydrogen, occupying a volume of 1.18 L at 894.6 mmHg and 44.1oC, it is found that the total mass of the sample is 1.25 g. What is the partial pressure of argon?

A 18.8-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 13.5 L when measured at 759.1 mmHg and 25.6oC. What is the mole fraction of carbon dioxide in this mixture?

Answer 1

(1)

Use equation:

PV = nRT

(894.6/760)*1.18 = n*0.0821*(273+44.1)

Solving we get:

Total moles, n = 0.053

Assume moles of argon as 'x' and that of hydrogen as 'y'

So,

40*x + 2*y = 1.25 ---(1)

x + y = 0.053 ---(2)

Solving (1) and (2) we get:

x = 0.03

So,

Mole fraction of Ar = X_Ar = 0.03/0.053 = 0.57

So,

Partial pressure of Ar = 894.6*X_Ar = 894.6*0.57 = 509.92 mm Hg

(2)

Use equation:

PV = nRT

(759.1/760)*13.5 = n*0.0821*(273+25.6)

Solving we get:

Total moles, n = 0.55

Assume moles of CO₂ as 'x' and that of N₂ as 'y'

So,

44*x + 28*y = 18.8 ---(1)

x + y = 0.55 ---(2)

Solving (1) and (2) we get:

x = 0.212

So,

Mole fraction of CO₂ = X_CO2 = 0.212/0.55 = 0.39