Respuesta :

Answer:

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 .