What volume of 0.125 M nitric acid can be prepared from 0.250 L of 1.88 M nitric acid?

1. Rearrange the dilution equation to V2 = C1V1 / C2. 2. Substituting gives V2 = 1.88 M x 0.250 L / 0.125 M. 3. The numerator is 0.470 mol and dividing by 0.125 M gives 3.76 L. 4. The concentration drops by a factor of about 15, so the volume must grow by about the same factor. 5. The value 0.470 L is the numerator alone, reported without the final division. _Source: OpenStax Chemistry (CC BY 4.0), Ch 3 "Composition of Substances and Solutions", section 3.3 Molarity_