Question

A balloon contains 475 cm of air at 798 mmHg and 26.4 °C. How many grams of oxygen are in the balloon? (Air is 21.0 mol % oxygen.)

Answer 1

Temperature = 26.4^oC (299.4 K)

Pressure = 798 mmHg (1.05 atm)

Volume = 475 cm3 (0.475 L)

R = 0.082057 atm L mol^-1K^-1

So, PV= nRT

1.05*0.475= n *0.082057* 299.4

Moles of air = 0.49875/24.5678

Moles of air = 0.0203

Moles of oxygen = 0.21* moles of air

Moles of oxygen = 0.21*0.0203

Moles of oxygen= 0.004263

So, mass of oxygen = moles of oxygen* molecular weight of oxygen

Mass of oxygen = 0.004263*1 6

Mass of oxygen in balloon = 0.068208g