Question

calculate the solubility of oxygen in water at 1.50 atm and 5°C if the solubility is 0.032 g oxygen per liter of water at 1.20 atm and 5°C

Answer 1

Solubility = mass/molar mass × 1000/vol

Volume is known

But PV = nRT

n= 0.032g/32gmol^-1 = 0.001mol

V = nRT/ P = 0.001mol × 0.08206atmkmol^-1 × 278k / 1.20atm

V=0.019L

Solubility = mole × 1000/vol

= 0.001mol × 1000/0.019L

= 52.63mol/L