Calculate the mass ( g ) of the solute ethanol and the mass ( kg ) of the solvent CHCl3 that should be combined to prepare 9900 g of a solution that is 14.5 m ethanol.

Question

extra given info from problem:
molar mass (g/mol):
C2H5OH 46.07
CHCl3 119.38

density (kg/L):
C2H5OH 0.7893
CHCl3 1.483

Answer 1

Check my work. It's late.

You want 9900 grams of 14.5 m solution of EtOH.
Suppose we wanted to prepare a 14.5 M solution the usual way. We would weigh out 145 moles EtOH and 10,000 g water (10 Kg). (Note that 145 moles/10 kg solvent = 14.5 m.) How much would that SOLUTION weigh? It will have a mass of 10,000 g for the water + the mass of 145 moles EtOH. 145 mols = 145 x 46.07 = 6680.15 g ( I know that's too many places but we can round at the end.)
The total mass of the solution will be 16,680.15 grams.

Now to the solution we want to prepare. We want the SOLUTION to be 9900 grams. So how many mols will be in that mass?
6680.15 x (9900/16680.15) = 3964.8 g EtOH. How much water? That will be
9900-3964.8 = 5935.2 g.
Will that do it?
3964.8 g EtOH = 3964.8/46.07 = 86.06 moles. m = 86.06 moles EtOH/5.9352 kg water = 14.4999 which rounds to 14.5 m.