Question

Calculate the amount of benzoic acid (C₆HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.

Answer 1

0.15 M solution means that 0.15 moles of C₆H₅COOH is present in 1L = 1000 mL of the solution 

Molar mass of C₆H₆COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol^-1 

Thus, 1000 mL of solution contains benzoic acid = 18.3 g

250 mL of solution will contain benzoic acid = \(\frac{18.3}{1000}\) x 250 = 4.575 g