An aqueous solution containing 15.0 mass% methanol (CH3OH; MM = 32.0 g/mol). Calculate the molality of the solution.

Answer 1

To find the molality of the solution, we need to determine the moles of methanol present in the solution.

Given:
Mass% of methanol = 15%
Molar mass of methanol (CH3OH) = 32.0 g/mol

Let's assume we have 100 grams of the solution.

The mass of methanol in the solution = 15% of 100 g = 15 g

Now, let's calculate the moles of methanol:
moles of CH3OH = mass of CH3OH / molar mass of CH3OH
moles of CH3OH = 15 g / 32.0 g/mol
moles of CH3OH = 0.46875 mol

Next, we need to calculate the mass of the solvent (water) in the solution.

mass of water = mass of solution - mass of CH3OH
mass of water = 100 g - 15 g
mass of water = 85 g

Now, we can calculate the molality:

molality = moles of solute / mass of solvent (in kg)
molality = 0.46875 mol / 0.085 kg
molality = 5.5147 mol/kg

Therefore, the molality of the solution is approximately 5.5147 mol/kg.