# "Chemistry" algebra

• Aug 28th 2010, 06:09 PM
WhoCares357
"Chemistry" algebra
I have this problem for homework:
Quote:

How many mL of 1M acetic acid (C2H4O2) 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.
It's a chemistry problem, but it's really just basic algebra. The different units represent the different variables.

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 (H2O); 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.
• Aug 29th 2010, 12:50 AM
CaptainBlack
Quote:

Originally Posted by WhoCares357
I have this problem for homework: It's a chemistry problem, but it's really just basic algebra. The different units represent the different variables.

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 (H2O); 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.

The concentration of molar acetic acid is ~=6%, and of 0.2 molar is ~=1.2% (by mass)

How much of each must be mixed to give 100mL of 5% solution?

CB
• Aug 29th 2010, 12:50 AM
CaptainBlack
Quote:

Originally Posted by WhoCares357
I have this problem for homework: It's a chemistry problem, but it's really just basic algebra. The different units represent the different variables.

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 (H2O); 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.

The concentration of molar acetic acid is ~=6%, and of 0.2 molar is ~=1.2% (by mass)

How much of each must be mixed to give 100mL of 5% solution?

CB
• Aug 29th 2010, 06:19 AM
WhoCares357
Quote:

Originally Posted by CaptainBlack
The concentration of molar acetic acid is ~=6%, and of 0.2 molar is ~=1.2% (by mass)

How much of each must be mixed to give 100mL of 5% solution?

CB

Edit: I think I figured out how you got the concentrations. I used the given volume for both formulas (molarity and molality) and got ~6g and ~1.14g.

I don't understand what I have to do with them now, though. Do I just add them?