Question

If the molarity of a solution of potassium carbonate is 0.567 m, what is the weight percent?

A) 7.27%

B) 35.3%

C) 18.4%

D) 3.91%

Answer 1

Note: There is a typo, molarity is modality.

Given, Molality of solution = 0.567 m = 0.567 mol/kg

Let, mole of potassium carbonate (K₂CO₃) = 0.567 mol

Mass of solvent (H₂O) = 1 kg = 1000 kg

Molar mass of K₂CO₃ = 138.205 g/mol

Mass of K₂CO₃ = Moles of K₂CO₃ x Molar mass of K₂CO₃

Mass of K₂CO₃ = 0.567 mol x 138.205 g/mol = 78.4 g

Mass of solution = mass of solute + mass of solvent

Mass of solution = 78.4 g + 1000 g = 1078.4 g

Weight percent =[(mass of solute)/(mass of solution)]x100 = [(78.4 g)/(1078.4)]x100 = 7.27%

Hence,

Answer: (A) 7.27%

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