Question

A sample of O2 is collected over water at 22.0 °C. The volume of the sample is 150.0 mL and the total pressure is 752 torr. What is the mass of O2 present? (The vapor pressure of water at 22 °C is 19.8 torr)

Answer 1

Given : Total pressure = 752 torr

vapor pressure of water = 19.8 torr

pressure of dry O₂ = (Total pressure) - (vapor pressure of water)

pressure of dry O₂ = (752 torr) - (19.8 torr)

pressure of dry O₂ = 732.2 torr

pressure of dry O₂ = 732.2 torr * (1 atm / 760 torr)

pressure of dry O₂ = 0.963 atm

According to ideal gas law,

moles of O₂ = [(pressure of dry O₂) * (volume of O₂)] / [(R) * (T)]

where volume of O₂ = 150.0 mL = 0.1500 L

R = gas constant = 0.0821 L-atm/mol-K

T = absolute temperature = (22 + 273) K = 295 K

Substituting the values,

moles of O₂ = [(0.963 atm) * (0.150 L)] / [(0.0821 L-atm/mol-K) * (295 K)]

moles of O₂ = 5.97 x 10^-3 mol

mass of O₂ = (moles of O₂) * (molar mass of O₂)

mass of O₂ = (5.97 x 10^-3 mol) * (32.0 g/mol)

mass of O₂ = 0.191 g