Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing 250 ml of 0.15 M solution in methanol.

Explanation

Molarity `= ("Moles of solute")/("Vol. of solution(in mL) ") xx 1000`
`therefore ` Moles of benzoic acid ` = (0.15 xx 250)/(1000) = 0.0375`
Molecular mass of benzoic acid `(C_(6)H_(5)C OOH)`
`= 7 xx 12+ 6 xx 1 2 xx 16`
` = 122`
Amount of benzoic acid ` = 0.375 xx 122` = 4.575g