Question

Calculate the mole fraction of a solution containing 2.6 grams of sodium sulfate in 20 grams of water

Answer 1

$n(Na_2SO_4)=2.6\ g\times\frac{1\ mol}{142.04\ g}=0.0183\ mol$

$n(H_2O)=20\ g\times\frac{1\ mol}{18.015\ g}=1.11\ mol$

$\chi(Na_2SO_4)=\frac{0.0183\ mol}{0.0183\ mol+1.11\ mol}=0.0162$

$\chi(H_2O)=\frac{1.11\ mol}{0.0183\ mol+1.11\ mol}=0.984$

Answer: Na₂SO₄ - 0.0162; H₂O - 0.984