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Question

A 2.96 L2.96 L aqueous solution of KOHKOH contains 181 g181 g of KOHKOH. The solution has a density of 1.29 g/mL1.29 g/mL.

Calculate the molarity (𝑀M), molality (𝑚m), and mass percent concentration of the solution.

Molarity=?

Molality=?

Mass Percent=?

Answer 1

Molality (m) = moles solute / kg solvent

moles solute = 181 g KOH x 1 mol KOH / 56.1 g = 3.23 mols KOH

kg solvent: 2.96 L x 1000 ml / L x 1.29 g / ml = 3818 g total mass of solution

mass of H2O = 3818 g - 181 g = 3637 g = 3.637 kg

m = 3.23 mols / 3.64 kg = 0.887 m

Molarity = moles solute / L solution

mols solute = 3.23 mols KOH (see above calculation)

Liters of solution = 2.96 L

M = 3.23 mol / 2.96 L = 1.09 M

Mass percent = grams KOH / grams solution (x100%)

grams KOH = 181 g

grams of solution = 3818 g (see above calculations)

mass % = 181 g / 3818 g (x100%) = 4.74%