What is the volume of solution prepared when 6g of acetic acid is dissolved to prepare 0.25M solution
mols acetic acid = grams/molar mass = ?
Then M acetic acid = mols/L solution. You know M and mols, solve for L.
Hilarious idiot answer u got there. 🙄
To find the volume of the solution, we need to use the equation:
Molarity (M) = moles of solute ÷ volume of solution (in liters)
First, we need to find the number of moles of acetic acid using its molar mass.
The molar mass of acetic acid (CH3COOH) is calculated as:
C: 12.01 g/mol
H: 1.008 g/mol (3 atoms)
O: 16.00 g/mol
Adding them up: 12.01 + (1.008 x 3) + 16.00 = 60.05 g/mol
Now, we can find the number of moles by dividing the given mass by the molar mass.
moles of acetic acid = mass ÷ molar mass
moles of acetic acid = 6 g ÷ 60.05 g/mol ≈ 0.09997 mol (approximately 0.1 mol)
Now, we can use the molarity formula to find the volume of the solution.
0.25 M = 0.1 mol ÷ volume of solution (in liters)
Rearranging the equation to solve for the volume of the solution:
volume of solution (in liters) = 0.1 mol ÷ 0.25 M = 0.4 L
Therefore, the volume of the solution prepared when 6 g of acetic acid is dissolved to prepare a 0.25 M solution is 0.4 liters or 400 milliliters (mL).