Benzoic Acid Required for Preparing 250 ml of 0.15 m Solution in Methanol

To calculate the amount of benzoic acid required to prepare 250 mL of a 0.15 M solution in methanol.

Formula

The formula to calculate the mass of a solute is:

$$\text{Mass of solute (g)} = M \times V \times M_m$$

Where:

• $M$ = Molarity of the solution ($0.15 \, \text{mol/L}$)
• $V$ = Volume of the solution in liters ($250 \, \text{mL} = 0.250 \, \text{L}$)
• $M_m$ = Molar mass of benzoic acid ($C_6H_5COOH$).

Molar Mass of Benzoic Acid

The molar mass of benzoic acid is calculated as:

$$M_m = (6 \times 12.01) + (5 \times 1.008) + (1 \times 12.01) + (2 \times 16.00) + (1 \times 1.008) = 122.13 \, \text{g/mol}$$

Plug Values into the Formula

$$\text{Mass of solute} = 0.15 \, \text{mol/L} \times 0.250 \, \text{L} \times 122.13 \, \text{g/mol}$$

Perform the Calculation

$$\text{Mass of solute} = 4.58 \, \text{g}$$

Final Answer

You need 4.58 g of benzoic acid to prepare 250 mL of a 0.15 M solution in methanol.