200 g of CH3COOH are dissolved in enough water to make 1.5 L. What is the molarity of the acetic acid solution?
mols = grams/molar mass, then
M = mols/L soln.
To calculate the molarity of the acetic acid solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's find the number of moles of CH3COOH:
Given mass of CH3COOH = 200 g
Molar mass of CH3COOH = 60.05 g/mol
Number of moles = mass / molar mass
Number of moles = 200 g / 60.05 g/mol ≈ 3.329 moles
Next, let's calculate the volume of the acetic acid solution:
Given volume of solution = 1.5 L
Now we can substitute the values into the formula:
Molarity (M) = 3.329 moles / 1.5 L
Molarity ≈ 2.22 M
Therefore, the molarity of the acetic acid solution is approximately 2.22 M.
To find the molarity of the acetic acid solution, we need to use the formula:
Molarity (M) = Number of moles of solute / Volume of solution (in liters)
First, we need to determine the number of moles of acetic acid (CH3COOH). We can use the formula:
Number of moles = Mass of solute / Molar mass of solute
The molar mass of acetic acid (CH3COOH) is calculated as follows:
Molar mass = (Number of carbon atoms x Atomic mass of Carbon) + (Number of hydrogen atoms x Atomic mass of Hydrogen) + (Number of oxygen atoms x Atomic mass of Oxygen)
Let's calculate the molar mass of CH3COOH:
Molar mass = (1 x 12.01 g/mol) + (3 x 1.01 g/mol) + (2 x 16.00 g/mol) + (1 x 1.01 g/mol)
Molar mass = 60.05 g/mol
Now, let's calculate the number of moles of acetic acid:
Number of moles = 200 g / 60.05 g/mol
Number of moles = 3.33 mol
Now, we need to calculate the molarity using the formula mentioned earlier:
Molarity (M) = Number of moles of solute / Volume of solution (in liters)
Molarity = 3.33 mol / 1.5 L
Molarity = 2.22 M
So, the molarity of the acetic acid solution is 2.22 M.