Question

Calculate the amount of benzoic acid ( C6H5COOH) required for preparing 250 ml of 0.15 M solution
methanol

Answer 1

Answer:

4.575gm

Explanation:

.15M= .15 mols in 1000ml

hence in 250 ml .15/4=.0375 mol= .0375×122=4.575 gm benzoic acid

(m.w. of benzoic acid =122)