Question

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 201 torr ; O2, 153 torr; and He, 113 torr

What mass of each gas is present in a 1.00 −L sample of this mixture at 25.0 ∘C?

Enter your answers numerically separated by commas.

Answer 1

1 atm = 760 torr

For N2(g)

Using ideal gas equation

PV = nRT

201/760 * 1 = n * 0.0821 * (273+25)

n = 0.01081 moles

Mass of N2 present = (Number of moles) * (Molar mass) = (0.01081 mol) * (28 g/mol) = 0.303 grams

For O2(g)

Using ideal gas equation

PV = nRT

153/760 * 1 = n * 0.0821 * (273+25)

n = 0.008228 moles

Mass of O2 present = (Number of moles) * (Molar mass) = (0.008228 mol) * (28 g/mol) = 0.263 grams

For He(g)

Using ideal gas equation

PV = nRT

113/760 * 1 = n * 0.0821 * (273+25)

n = 0.006077 moles

Mass of He present = (Number of moles) * (Molar mass) = (0.006077 mol) * (4 g/mol) = 0.0243 grams

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