How many grams of glacial acetic acid should be used to make 250 mL of 2.00 M acetic acid? Show all your work.

To calculate the number of grams of glacial acetic acid needed to make 250 mL of 2.00 M acetic acid, we need to follow these steps:

Step 1: Convert the volume in milliliters to liters.
250 mL = 250/1000 = 0.25 L

Step 2: Use the formula for molarity to calculate the number of moles of acetic acid.
Molarity (M) = moles of solute / volume of solution in liters
2.00 M = moles of acetic acid / 0.25 L

Rearranging the formula, we get:
moles of acetic acid = Molarity (M) x volume of solution in liters
moles of acetic acid = 2.00 M x 0.25 L = 0.50 moles

Step 3: Use the molar mass of acetic acid to calculate the number of grams needed.
The molar mass of acetic acid (CH3COOH) is:
(1 x 12.01 g/mol) + (2 x 1.008 g/mol) + (4 x 16.00 g/mol) = 60.052 g/mol

The number of grams of acetic acid needed can be calculated using the formula:
grams = moles x molar mass
grams = 0.50 moles x 60.052 g/mol ≈ 30.03 g

Therefore, you would need approximately 30.03 grams of glacial acetic acid to make 250 mL of 2.00 M acetic acid.

To calculate the number of grams of glacial acetic acid needed to make a specific concentration of acetic acid solution, we need to know the molecular weight of acetic acid. The chemical formula for acetic acid is CH3COOH, and its molecular weight is 60.05 g/mol.

To calculate the number of moles of acetic acid required, we can use the formula:

moles of solute (acetic acid) = concentration (M) × volume (L)

Given that the desired concentration of acetic acid is 2.00 M and the volume is 0.250 L (since 250 mL is equivalent to 0.250 L), we can substitute these values into the formula:

moles of acetic acid = 2.00 M × 0.250 L = 0.500 mol

Since we now know the moles of acetic acid required, we can calculate the mass (in grams) using the formula:

mass (g) = moles × molecular weight

Substituting the values, we get:

mass of acetic acid = 0.500 mol × 60.05 g/mol = 30.025 g

Therefore, you would need approximately 30.025 grams of glacial acetic acid to make 250 mL of 2.00 M acetic acid solution.

Mol wgt. of acetic acid is 60grams in 1litre is 1M for 2M dissolve 120gram of glacial acetic acid in 1litre toget d required volume of 250ml divide1litre by 4 do the same to 120grams thus weigh only 30g of glacial acetic acid in make up to250ml mark this will give the required 2M