Question

How many grams of CH3OH (MW = 32.04) are in 900 g of a 1.5 m aqueous solution?

I got 43.3 g but the answer is 41.3 g. I don't know where I went wrong.

I converted 1.5 m (molality) into .0015 mol/g multiplied by 900 g and multiplied again by 32.04 g/mol.

Answer 1

M = molarity = (mol solute)/(volume of entire solution)
m = molality = (mol solute)/(mass of just the solvent)

Note that that says solvent not the entire solution, this is where your error is coming from. However, correcting it is not that simple. You need to figure out the mass of solvent in the solution, but they don't give you the density. So you need to take a round about method.

• a 1.5m solution = 1.5mols solute for every 1kg of solvent.

We need to find the mass of the solution. Luckily they gave you the molecular weight of the solute.

• 1.5mols CH3OH = 48.06g CH3OH
• 1000g solvent + 48.06g CH3OH = 1048.06g solution

So now we can show that for every 1048.06g solution there are 1.5mols of CH3OH. lets simplify that a bit.

• 1048.06g solution / 1.5mols CH3OH = 698.7g solution/mol CH3OH

You have 900g solution right?

• (900g solution)/(698.7g solution/mol CH3OH) = 1.29 mols CH3OH
• (1.29 mols CH3OH)*(32.04g/mol CH3OH) = 41.3 g CH3OH