I need help with calculating percent mass and molarity in the questions below.
Vinegar is made by adding 33 g of acetic acid and 625 g water. What is the percent by mass of acetic acid in the solution?
Assuming the density of vinegar to be 1.0 g/mL, what is the molarity? The molar mass of acetic acid is 60.05 g/mol.
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The percent mass of acetic acid in vinegar is found by dividing the mass of acetic acid by the total mass of the solution (acetic acid plus water) and multiplying by 100:
33 g aa/((33 g aa + 625 g water)*100=5.02% mass acetic acid
For the molarity part you had a typo in your question that I corrected. You left off the "6" for the molar mass of acetic acid. Molarity is defined as moles solute per liter solution so we need to find the total number of moles of acetic acid and divide it by the total volume of the solution. The moles of acetic acid are found by dividing the mass of acetic acid by its molar mass as given above:
33 g aa*(1 mole aa/60.05 g aa)=0.55 moles acetic acid
The total volume of the solution is found by dividing the total mass of the solution (acetic acid plus water) by the density of vinegar as given above:
658 g vinegar*(1 mL/1 g)=658 mL vingar (since the density is 1 then the mass is the same as the volume).
Now divide the moles of acetic acid by the volume of vinegar (in liters) to get the molarity:
0.55 moles aa/0.658 L=0.84 M
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