Assuming the density of vinegar is 1.0 g/mL ' what is the molarity of vinegar? (use the percent by mass of acetic acid to vinegar which is 5%.) The molar mass of acetic acid is 60.05 g/mol?
%w/w = grams solute/100 g solution.
%w/w = 5 grams CH3COOH/100 mL soln.
moles CH3COOH = 5/60.05 = ?? moles/100 mL.
Then M = moles/L soln.
Household vinegar is aquous, 5% acetic acid HC2H3O2 solution by mass. The molarity is: 5% = 5g
1mol HC2H3O2 = 60.05g HC2H3O2
5g/60.05g = 0.08326394671 mol
therefore the molarity is 0.08326394671 mol/0.01L = 0.84mol/L
Well, first of all, let me just say that vinegar is quite a pickle of a substance! Now, to the math. We know that the percent by mass of acetic acid in vinegar is 5%, and the molar mass of acetic acid is 60.05 g/mol.
To find the molarity of vinegar, we need to convert the percent by mass to grams of acetic acid per liter of solution. So, assuming we have 1 liter of vinegar (which is about the same as 1 kilogram), we can calculate it like this:
5% of 1000 g (1 liter) = 50 g of acetic acid
Now that we have the mass of acetic acid, we can convert it to moles using the molar mass:
50 g / 60.05 g/mol = approximately 0.8336 mol
Finally, we divide the moles of acetic acid by the volume of the solution in liters:
0.8336 mol / 1 L = approximately 0.8336 M
So, the molarity of vinegar is about 0.8336 M. Just be careful not to accidentally pickle yourself in the process!
To find the molarity of vinegar, we need to calculate the number of moles of acetic acid in 1 L of vinegar.
Step 1: Calculate the mass of acetic acid in 1 L of vinegar.
Given that the density of vinegar is 1.0 g/mL, we can convert 1 L to grams by multiplying it by the density:
1 L * 1.0 g/mL = 1000 g
Since acetic acid constitutes 5% of vinegar, we can calculate the mass of acetic acid in 1 L of vinegar:
1000 g * 0.05 = 50 g
Step 2: Calculate the number of moles of acetic acid.
The molar mass of acetic acid is given as 60.05 g/mol. To find the number of moles, divide the mass by the molar mass:
50 g / 60.05 g/mol ≈ 0.8337 mol
Step 3: Calculate the molarity.
Molarity is defined as the number of moles of solute per liter of solution.
Since 0.8337 moles of acetic acid are in 1 L of vinegar, the molarity is:
0.8337 mol/L or approximately 0.8337 M
Therefore, the molarity of vinegar is approximately 0.8337 M.
To find the molarity of vinegar, we need to use the given information about the density and the percent by mass of acetic acid.
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. In this case, the solute is acetic acid, and the solvent is vinegar.
First, we need to determine the mass of acetic acid in 1 liter of vinegar. Since we know that the density of vinegar is 1.0 g/mL, this means that 1 liter of vinegar weighs 1000 grams.
Next, we need to calculate the mass of acetic acid in 1000 grams of vinegar. We can do this by multiplying the mass of vinegar by the percentage of acetic acid. In this case, 5% of acetic acid means 5 grams of acetic acid in every 100 grams of vinegar.
Therefore, the mass of acetic acid in 1000 grams (1 liter) of vinegar is:
(5 grams / 100 grams) x 1000 grams = 50 grams
Now that we know the mass of acetic acid in 1 liter of vinegar, we can calculate the number of moles of acetic acid using the molar mass of acetic acid.
The molar mass of acetic acid is given as 60.05 g/mol. So, we divide the mass of acetic acid by its molar mass to get the number of moles:
50 grams / 60.05 g/mol = 0.833 mol
Finally, we can calculate the molarity by dividing the moles of acetic acid by the volume of vinegar solution in liters:
Molarity (M) = moles of solute / liters of solution
Molarity = 0.833 mol / 1 liter = 0.833 M
Therefore, the molarity of vinegar is 0.833 M.