Final answer:
To prepare a 0.15 molar solution of benzoic acid in methanol, you would need approximately 4.59 grams of benzoic acid.
Step-by-step explanation:
To calculate the amount of benzoic acid (C₆HCOOH) required to prepare a 0.15 molar solution in methanol, we can use the formula:
Amount (in moles) = Molarity × Volume (in liters)
First, we need to convert the volume from milliliters to liters. 250 ml is equal to 0.25 liters. Substituting the values into the formula, we get:
Amount (in moles) = 0.15 M × 0.25 L = 0.0375 moles
To calculate the amount in grams, we can use the molar mass of benzoic acid, which is 122.13 g/mol. Substituting again, we get:
Mass (in grams) = Amount (in moles) × Molar mass = 0.0375 moles × 122.13 g/mol = 4.59 grams
Therefore, you would need approximately 4.59 grams of benzoic acid to prepare a 250 ml 0.15 molar solution in methanol.