I have this problem for homework:

How many mL of 1M acetic acid (C_{2}H_{4}O_{2}) and .2 molal acetic acid need to be added to make 100 mL of vinegar (5% acetic acid)? Assume 1g/mL density of all solutions.

I started by listing everything that is given: density = 1g/mL; 1M solute = 1 mol solute/1 L solution; .2 molal solute = .2mol solute/1kg solvent (H

_{2}O); 60.052 g/mol solute; 18.000 g/mol solvent;

**.1L total solution**. I am looking for the mL of solute in .1L of solution.

My first step was to use the Molarity to figure out the number of moles of the solute: 1M solute = .1 mol solute/.1L solution.

Then I used the number of moles to find the number of grams in the solution: .1mol x 60.052g/mol = 6.0052g solute.

Finally I converted the mass to volume using density: 6.0052g/(1g/mL) = 6.0052mL.

However I am pretty sure that this answer is wrong. When I insert the .1 mol of solute into the molality formula I get a totally different answer.

.2 molal = .1mol/500g -> 500gx.05/.95 = 26.315g -> 26.315g/(1g/mL) = 26.315mL.

26.315mL=/= 6.0052mL

Can anyone show me where I went wrong. It's possible that I may have just overthought it, but the answer just doesn't seem right to me.

Thank you.