# Finding the molality of a solution when molarity adn desnity is given

Submitted by jkretsch on Thu, 12/08/2011 - 16:43

I need help trying to solve to problem:

The density of a 15.9 M aqueous CH3OH (methanol) solution is 0.808 g/mL. What is the molality of this solution?

How do I solve this?

This is a pretty long problem. It comes down to carefully analysis of the units behind molarity (M) and molality.

Molarity is moles of solute / Liters of solution

molality is moles of solute / kg of solvent

So the easiest way is to assume you have 1.0 L of a solution.

1.0L x 15.9 M = 15.9 moles of solute

Use the molar mass of the solute methanol and you can fine the mass of the solute.

15.9 moles of solute x   32   g / mol = 508.8 grams

Use the assumed 1.0 L of solution and density to find the mass of the solution

1000 mL x 0.808 g/mL = 808 g of solution

Since a solution is made up of a solute and solvent

mass of the solution = mass of solute + mass of solvent

mass of solvent = 808 - 508.8 = 299.2 g

now plug into equation for molarity

molality  = 15.9 moles / 0.2992 kg = 53.1 molal