You are given 350 mL a stock solution of 8.32E-03 M benzoic acid, MW = 122.125 g/mol.

If the actual density should be 1.05 g/mL. What is your percent error?

an approximate error can be by assuming there is 350grams of water and .350*8.32E-3*122.125, then dividing it by .350L. That will give an approximate density. I am not suggesting this is a good way here, but the data is insufficient.

To find the percent error, we need to calculate the expected mass of the stock solution based on its volume and density, and then compare it to the actual mass given the molar mass of benzoic acid.

First, we need to calculate the expected mass of the stock solution. The density is given as 1.05 g/mL, and the volume is given as 350 mL. We can use the formula:

Mass = Density * Volume

Substituting the values:

Mass = 1.05 g/mL * 350 mL
Mass = 367.5 g

Next, we need to determine the moles of benzoic acid in the stock solution. The molar mass of benzoic acid is given as 122.125 g/mol. We can use the formula:

Moles = Mass / Molar mass

Substituting the values:

Moles = 367.5 g / 122.125 g/mol
Moles = 3.010 mol

Now we can calculate the expected concentration by dividing the moles of benzoic acid by the volume:

Concentration = Moles / Volume

Substituting the values:

Concentration = 3.010 mol / 350 mL
Concentration = 8.6E-03 M

Finally, we can calculate the percent error by comparing the expected concentration to the given concentration of 8.32E-03 M:

Percent Error = |Expected Concentration - Given Concentration| / Given Concentration * 100

Substituting the values:

Percent Error = |8.6E-03 M - 8.32E-03 M| / 8.32E-03 M * 100
Percent Error = |0.28E-03 M| / 8.32E-03 M * 100
Percent Error = 0.28E-03 M / 8.32E-03 M * 100
Percent Error = 3.36%

Therefore, the percent error is 3.36%.