If 10.0 g of sodium sulfite is dissolved in 1.00 kg of water to yield a concentration of 1.06 g/mL, calculate the concentration of the solution

Question

using molarity, molality, percent by mass, mole fraction

Answer 1

sodium sulfite = Na₂SO₃; molar mass = 126 g/mol

molality (m) = moles solute/kg solvent = 10.0 g x 1 mol/126 g /1.00 kg = 0.0794 m

percent by mass = mass Na₂SO₃/Total mass of solution (x100%)

mass Na₂SO₃ = 10.0 g

total mass = 10.0 g + 1.0 kg x 1000 g/kg = 1010 g

% by mass = 10.0 g / 1010 g (x100%) = 0.990%

mole fraction = moles Na₂SO₃/total moles

moles Na₂SO₃ = 10.0 g x 1 mol/126 g = 0.0794 moles

moles H₂O = 1.00 kg x 1000 g/kg x 1 mol/18 g = 55.56 moles

total moles = 55.64 moles

mole fraction Na₂SO₃ = 0.0794/55.64 = 0.00143

molarity (M) = moles Na₂SO₃/liter solution:

assuming that the question meant to state that the density of solution is 1.06 g/ml (not the concentration)...

1010 g soln x 1 ml/1.06 g = 952.8 mls of solution

0.0794 mol Na₂SO₃/952.8 mls x 1000 ml/L = 0.0833 mol/L = 0.0833 M