A person accidentally swallows a drop of liquid oxygen, O2 (l), which has a density of 1.149 g/mL. Assuming the drop has a volume of 0.050 mL, what volume (mL) of gas will be produced in the person’s stomach at body temperature (37 °C) and a pressure of 1.0 atm?

Answer 1

mass = volume x density = 0.05 mL x 1.149 g/mL = ? g O2.
n = mols O2 = grams/molar mass = ?

Then PV = nRT. Remember T must be in kelvin. V will be in L so convert to mL. Post your work if you get stuck.

Answer 2

The answer should be 0.0457 L