Answer to Question #139812 in General Chemistry for Harvey Jed A. Iman

Question

What is the molality of 60.5% by mass of nitric acid (HNO3) solution?
Calculate the mole fraction of HCl and H2O in a 20% (w/w) aqueous HCl solution.

Answer 1

M (HNO₃) = 63.01 g/mol

If 100 g of sln given, then m (HNO₃) = 60.5 g, m (H₂O) = 100 - 60.5 = 39.5 g (0.0395 kg).

moles HNO₃ = 60.5 / 63.01 = 0.9602 mol

molality = 0.9602 mol / 0.0395 kg = 24.3 m HNO₃

20% (w/w) aqueous HCl solution means 20 g HCl + 80 g of water = 100 g of solution

M (HCl) = 36.46 g/mol

M (H₂O) = 18.02 g/mol

20 / 36.46 = 0.5485 mol HCl

80 / 18.02 = 4.4395 mol H₂O

Mole fraction = 0.5485 / (0.5485 + 4.4395) = 0.11 or x100% = 11%