Question

A sample of oil (density: 0.89g/mL) was found to have dioxin contamination of 2ppm. How many milliliters of the oil would contain exactly 0.01 gram of dioxin?

a) 2.25 mL

b) 5.62 mL

c) 11.24 mL

d) 22.47 mL

Answer 1

To determine how many milliliters of oil with a density of 0.89 g/mL and dioxin contamination of 2 ppm would contain 0.01 gram of dioxin, one must calculate the mass of dioxin per milliliter of oil and then divide 0.01 g by that concentration. The result is 5.62 mL of oil.

Step-by-step explanation:

To calculate how many milliliters of oil would contain exactly 0.01 gram of dioxin given a dioxin contamination of 2 ppm, we need to understand the concept of ppm (parts per million). Ppm is a way of expressing very dilute concentrations of substances. One ppm is equivalent to 1 milligram of a substance in 1 liter of liquid if the density is 1 g/mL. Since the density of the oil is 0.89 g/mL, we first need to convert the mass of dioxin to the equivalent volume in oil.

To find the mass of dioxin in grams per milliliter of oil, we multiply the density of the oil by the ppm value:

0.89 g/mL (oil density) × 2 ÷ 1,000,000 ppm = 1.78 × 10^-6 g/mL.

To get 0.01 g of dioxin, we divide the desired mass of dioxin by the concentration in g/mL:

0.01 g / (1.78 × 10^-6 g/mL) = 5,617.98 mL.

Thus, it would take 5.62 mL of oil to contain exactly 0.01 gram of dioxin, which corresponds to option b).