CHEMISTRY QUESTION NEED HELP!! The density of the acetic acid solution is 1.05 g/mL. Calculate the %(m/m) of the acetic acid solution (convert mol/L to %(m/m)).

Question

The density of the acetic acid solution is 1.05 g/mL. Calculate the %(m/m) of the acetic acid solution (convert mol/L to %(m/m)).

Volume of CH₃COOH = 0.1L

Titration Equation: CH3COOH(aq) + NaOH(aq) = CH3COOHNa(aq) + H2O(L)

Concentration of CH₃COOH = 0.243 mol/L

Answer 1

For %m/m, you need to divide the mass of the acetic acid / mass of the solution

1st you need to take the concentration of CH₃COOH to moles (think about units) If you have 0.243 mol/L and a volume of 0.1L. Would you multiply or divide the numbers to get rid of the L and leave only mols?

2nd Once you have the moles, now you need to convert the moles to grams using molecular mass of acetic acid

3rd Now you need to get the mass of the solution. You start off with 0.1L. Using the density you can get the mass of the solution. Density is g/mL. Convert the 0.1L to mL. Then multiply this number by the density to get the volume to cancel out and leave only the grams.

4th Now you have all the information. Take the mass from step 2 and divide by the mass from step 3. Multiply by 100 to get a percentage.