Question

63 g HNO₃ is in the dilute solution of 200 ml HNO₃ (Nitric acid). Find the molarity.

Answer 1

GMM of HNO₃ = 1 + 14 + 3 x 16 = 63 g

n = \(\frac{6.3}{63}\) = 0.1;V = 200ml = 0.2l;

M = \(\frac{n}{v}\) = \(\frac{0.1}{0.2}\) = 0.5 M