What is the molarity of CH3COOH in vinegar containing 4.0% CH3COOH by mass and having a density of 1.02 g/ml?
density of 1.02 g/mL x 1000 mL = 1020 g/L. grams CH3COOH in that 1000 mL is 1020 x 0.04 = 40.8 grams.
mols in that 40.8 g = 40.8/molar mass CH3COOH = ?
Since that is the mols/L then that is the molarity.
Well, let's see if I can whip up a "saucey" answer for you! To find the molarity of CH3COOH in vinegar, we'll need to do a little bit of chemistry math.
First, we need to convert the percentage of CH3COOH by mass to grams. Assuming we have 100 grams of vinegar, 4.0% of that would be 4.0 grams of CH3COOH.
Next, we need to find the volume of vinegar. We can use the density of vinegar (1.02 g/ml) to determine this. Since we have 100 grams of vinegar, we can divide that by the density to get the volume:
100 g / 1.02 g/ml ≈ 98.04 ml
Now, to find the molarity, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Since the molar mass of CH3COOH is about 60 g/mol, we can divide the mass of CH3COOH by the molar mass to get the moles:
4.0 g / 60 g/mol ≈ 0.067 mol
Finally, we can divide the moles of CH3COOH by the volume of vinegar in liters:
0.067 mol / 0.09804 L ≈ 0.68 M
So, the molarity of CH3COOH in the vinegar is approximately 0.68 M. It's interesting how something as simple as vinegar can have such a complex personality, isn't it? Now, go forth and dazzle your chemistry class with your newfound knowledge!
To determine the molarity of CH3COOH in vinegar, we need to follow these steps:
Step 1: Convert the percentage of CH3COOH by mass to grams.
Given that the vinegar contains 4.0% CH3COOH by mass, we can assume we have 100g of the vinegar. Therefore, the mass of CH3COOH in the vinegar is:
4.0g CH3COOH = (4.0/100) x 100g vinegar = 4.0 g
Step 2: Calculate the volume of the vinegar.
We are given that the vinegar has a density of 1.02 g/ml, and we have 100g of vinegar. Using the density formula:
Density = Mass / Volume
Plugging in the given values:
1.02 g/ml = 100g / Volume
Rearranging the equation to solve for volume:
Volume = 100g / 1.02 g/ml
Volume = 98.04 ml
Step 3: Calculate the number of moles of CH3COOH.
Given the molar mass of CH3COOH is 60.05 g/mol, we can calculate the number of moles using the formula:
Moles = Mass / Molar mass
Plugging in the given values:
Moles = 4.0g / 60.05 g/mol
Moles = 0.0666 mol
Step 4: Calculate the molarity of CH3COOH.
Molarity is defined as moles per liter of solution. We need to convert the volume from milliliters to liters:
Volume = 98.04 ml x (1 L / 1000 ml)
Volume = 0.09804 L
Now we can calculate the molarity:
Molarity = Moles / Volume
Plugging in the values:
Molarity = 0.0666 mol / 0.09804 L
Molarity = 0.678 M
Therefore, the molarity of CH3COOH in the vinegar is approximately 0.678 M.
To find the molarity of CH3COOH in vinegar, we will need to use the given information about the mass percentage of CH3COOH and the density of vinegar.
First, let's calculate the mass of CH3COOH in the vinegar.
Given:
Mass percentage of CH3COOH in vinegar = 4.0%
Density of vinegar = 1.02 g/ml
We can assume we have 100 g of vinegar.
Mass of CH3COOH = (Mass percentage of CH3COOH / 100) * Mass of vinegar
Mass of CH3COOH = (4.0 / 100) * 100 g
Mass of CH3COOH = 4.0 g
Next, we need to convert the mass of CH3COOH to moles.
Molar mass of CH3COOH = 12.01 g/mol + 1.01 g/mol * 3 + 16.00 g/mol + 1.01 g/mol = 60.05 g/mol
Number of moles of CH3COOH = Mass of CH3COOH / Molar mass of CH3COOH
Number of moles of CH3COOH = 4.0 g / 60.05 g/mol
Number of moles of CH3COOH ≈ 0.0666 mol
Finally, we can calculate the molarity of CH3COOH.
Molarity = Number of moles of CH3COOH / Volume of solution in liters
To find the volume of the solution, we need to use the density. Since we know that the density is 1.02 g/ml, we can calculate the volume using the formula:
Volume of solution = Mass of solution / Density of solution
Given:
Mass of solution = 100 g (assumed)
Density of solution = 1.02 g/ml
Volume of solution = 100 g / 1.02 g/ml
Volume of solution ≈ 98.04 ml ≈ 0.09804 L
Now we can calculate the molarity:
Molecular weight of CH3COOH = 60.05 g/mol.
Volume of solution = 0.09804 L.
Molarity = Number of moles of CH3COOH / Volume of solution
Molarity = 0.0666 mol / 0.09804 L
Molarity ≈ 0.679 M
Therefore, the molarity of CH3COOH in vinegar is approximately 0.679 M.