Question

A 0.550 m aqueous solution of KBr has a total mass of 77.0 g. What masses of solute and solvent are present?

gKBr=

gH2O=

Answer 1

Mass of KBr + mass of water in g = 77 g

Molality of aqueous solution (KBr + water) = 0.550

Moles of KBr / mass of water in kg = 0.550

(Mass of KBr)/(molar mass of KBr x mass of water in kg) = 0.550

Mass of KBr = 0.550 mol/kg x 119 g/mol x (mass of water in kg)

1000 x (77g - mass of water in g) = 65.45 x mass of water in g

77000 g = 65.45 x mass of water in g + 1000 x Mass of water in g

Mass of water in g = 77000/(1000+65.45) = 72.27 g

Mass of KBr = 77 - 72.27 = 4.73 g