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# How can I find the molality of a solution that contains 1.80 g of glucose (M. W. 180) and 10.0 g of water?

357 Points
12 years ago

Hi

1. Calculate the number of moles of solute. In this problem, it's 1.80 g glucose × (1 mol glucose / 180.0 g glucose) = 0.0100 mol glucose.
2. Calculate the number of kilograms of solvent. In this case, it's 10.0 g water × (1 kg water / 1000 g water) = 0.0100 kg water.
3. Divide moles of solute by kilograms of solvent This solution has a molality of 0.0100 mol glucose / 0.0100 kg water = 1.00 mol glucose / kg water. You'd write the concentration as 1.00 m glucose on the label.

But you probably want to know how to convert molarity into molality. If the solution is very dilute, you can treat molality and molarity as about the same thing, because 1 L of dilute solution contains about 1 kg of water. In concentrated solutions, molality and molarity are NOT equal, and you must know the density (or specific gravity) of the solution to convert one into the other. For example, to find the molality of a 3.00 M glucose solution with a density of 1.02 g/mL, follow this strategy:

1. Calculate the number of moles of solute. It's 3.00 mol glucose in 1 L of solution.
2. Calculate the number of kilograms of solvent. Find this as the difference between the mass of the solution and the mass of the solute.
* The mass of the solution is 1 L × (1000 mL / 1 L) × ( 1.02 g / mL) × (1 kg / 1000 g) = 1.02 kg.
* The mass of solute is 3.00 mol glucose × (180 g glucose / 1 mol glucose) × ( 1 kg / 1000 g ) = 0.54 kg.
* The mass of the solvent is 1.02 kg - 0.54 = 0.48 kg.
3. Divide moles of solute by kilograms of solvent. This solution has a molality of 3.00 mol glucose / 0.48 kg water = 6.25 m.