If 1.025 g of unknown solute is added to 4.568g of lauric acid and the sloution is found to have a molality of 1.81 m, How many moles of solute are in the solution?

Answer 1

molality = moles of solute / kg solvent

moles of solute = ?

kg of solvent = 4.568 g x 1 kg/1000 g = 0.004568 kg

molality = moles/kg = 1.81

1.81 moles/kg = x moles / 0.004568 kg

x = 0.00827 moles of solute are in the solution

The fact that 1.025 g is used does not enter into the calculation unless you wanted to determine the molar mass of the compound).