Calculation of Molarity of Ethanoic Acid in Benzene

Given Data:
- Mass of Ethanoic Acid (CH3COOH) = 2.5 grams
- Mass of Benzene = 75 grams

Formula to calculate Molarity:
Molarity (M) = (n/V)

Where,
- n = moles of solute
- V = volume of solution in liters

Step 1: Calculate the moles of Ethanoic Acid (CH3COOH)
To calculate the moles of Ethanoic Acid, we need to use the formula:

Moles (n) = Mass (m) / Molar mass (M)

The molar mass of Ethanoic Acid (CH3COOH) is calculated as follows:
C: 12.01 g/mol
H: 1.01 g/mol x 4 = 4.04 g/mol
O: 16.00 g/mol

Molar mass of CH3COOH = 12.01 + 4.04 + 16.00 = 32.05 g/mol

Now, we can calculate the moles of Ethanoic Acid:
n = 2.5 g / 32.05 g/mol
n = 0.078 moles

Step 2: Calculate the volume of the solution in liters
The volume of the solution can be calculated using the formula:

Volume (V) = Mass (m) / Density (D)

The density of Benzene is typically around 0.879 g/mL.

Volume of Benzene = 75 g / 0.879 g/mL
Volume of Benzene = 85.32 mL

As we need the volume in liters, we convert it:
Volume of Benzene = 85.32 mL / 1000 mL/L
Volume of Benzene = 0.08532 L

Step 3: Calculate the Molarity
Now, we can substitute the values into the molarity formula:

Molarity (M) = 0.078 moles / 0.08532 L
Molarity (M) = 0.913 M

Conclusion:
The molarity of 2.5 grams of Ethanoic Acid (CH3COOH) in 75 grams of Benzene is 0.913 M.

0.537M