Question

An certain compound has a solubility of 169 g/l in a solvent. If the mixture contains 50g of solute in a 300 cm^3 of solution, is the solution unsaturated? If yes, how many grams of solute should be added to make a saturated solution?

Answer 1

Proportion:

50 g — 300 cm3

x — 1000 cm3

x = 166.6 g/L (The solution is unsaturated)

Proportion:

169 g — 1000 cm3

y — 300 cm3

y = 50.7 g

∆m = 50.7 — 50 = 0.7 g (mass of solute that should be added to make a saturated solution)

Answer: 0.7 g