Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing 250 ml of 0.15 M solution in methanol.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
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.”`