Find the volume of a 2.5 M solution of sulfuric acid required to prepare 300.0 mL of 0.1 N solution of sulfuric acid

Answer 1

M = (1/2)*N = 1/2 * 0.1 = 0.05
mL1 x M1 = mL2 x M2
mL2 x 2.5 = 300*0.05