A solution contains 68.4 g of sucrose (C12H22O11) dissolved in 250 g of water. What
is the molality of the solution?
Molality of a solution is the moles of solute divided by the kilograms of solvent.
Molality = Moles of solute (mol) / Mass of solvent (kg)
solute = sucrose (C12H22O11)
solvent = water (H2O)
Mass of H2O = 250 g = 0.25 kg
Mass of C12H22O11 = 68.4 g
The molar mass of C12H22O11 is 342.3 g/mol
(68.4 g C12H22O11) × (1 mol C12H22O11 / 342.3 g C12H22O11) = 0.1998 mol = 0.2 mol C12H22O11
Moles of C12H22O11 = 0.2 mol
Molality of the solution = (0.2 mol) / (0.25 kg) = 0.8 mol/kg = 0.8 m
Molality of the solution = 0.8 m
Answer: The molality of the solution is 0.8 m