Calculate the amount of benzoic acid `(C_6H_3COOH)` required for preparing 250 mL of 0.15 M solution in methanol

Explanation

0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L, i.e., 1000 mL of the solution.
Molar mass of benzoic acid `(C_6H_5COOH) = 72 + 5 + 12 + 32 + 1 = 122 g "mol"^(-1)`
` therefore `0.15 mole of benzoic acid contains = 0.15 x 122 g = 18.3 g of benzoic acid
Thus, 1000 mL of the solution contain = 18.3 g benzoic acid
` therefore `250 mL of the solution will contain =` (18.3)/(1000) xx 250 = 4.575 g `benzoic acid