A 0.780 𝑚 aqueous solution of KBr has a total mass of 63.0 g. What masses of solute and solvent are present?

Answer 1

0.780 m = 0.780 moles KBr + 1000 g H₂O (by definition)

molar mass KBr = 119 g/mol

If we assume we used 1000 g H₂O (1 kg), then...

Mass KBr = 0.780 moles x 119 g /mol = 92.8 g KBr

In 63.0 g of solution we would have...

63.0 g soln x (92.8 g KBr / 92.8 g + 1000 g) = 63.0 x 0.0849 = 5.35 g KBr

63 g soln x (1000 g H₂O / 92.8 g + 1000 g) = 63.0 x 0.915 = 57.7 g H₂O