Calculate the amount of benzoic acid required for preparing 250mL of 0.15M solution in methanol.

Question

Philoid · Accepted Answer

Volume of the solution = 250mL = 0.25L

Let the no. Of moles of solute be n

Molarity = No. of moles of solute/volume of solution

⇒ 0.15 = n/0.25

⇒ n = 0.0375moles

Molar mass of C₆H₅OH = 6×12 + 5×1 + 16 + 1 = 94g

Moles = mass/molar mass

⇒ 0.0375 = m/94

Mass of benzoic acid required = 3.525g.