Question

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

Answer 1

Given
Molarity = 0.15 M
Volume V = 250 ml
Molecular weight of benzoic acid `(C_(6)H_(5)COOH)=122`
Molarity (M) `=("Weight")/(GMw)xx(1000)/(V(ml))`
`0.15=(W)/(122)xx(1000)/(250)`
`W=(122xx0.15)/(4)=4.575" gms."`