Answer:
Step-by-step explanation:
159 mL of .135 M benzoic acid will contain
.159 x .135 = .021465 moles of benzoic acid.
Molecular weight of benzoic acid = 122 gm
grams of .021465 moles = 122 x .021465 = 2.6 grams .
So 2.6 grams of benzoic acid will be required .