The problem involves calculating the maximum mass of chloroform (in micrograms, µg) that can be present in a given volume of groundwater while meeting the health standard set by the Minnesota Department of Health.

Information provided:

• Health risk limit for chloroform in groundwater: 60.0 µg/L
• Measured volume of the sample: 77.0 mL

Approach:

1. First, convert the volume from milliliters (mL) to liters (L), since the concentration is given in µg/L: $77.0 \, \text{mL} = 0.0770 \, \text{L}$

2. The concentration of chloroform is given as 60.0 µg/L. To find the maximum mass of chloroform in this sample, multiply the concentration by the volume: $\text{Mass of chloroform} = 60.0 \, \text{µg/L} \times 0.0770 \, \text{L} = 4.62 \, \text{µg}$

Thus, the maximum mass of chloroform that can be present in the 77.0 mL sample while meeting the health standard is 4.62 µg.

Would you like a more detailed explanation of any part of the calculation?

Here are five related questions to expand your understanding:

1. How would the result change if the concentration were halved?
2. What would happen if the volume were given in liters instead of milliliters?
3. How can significant figures affect the final answer in this kind of calculation?
4. How would you calculate the mass if the concentration were given in mg/L instead of µg/L?
5. What other factors might influence the safe concentration of a chemical in groundwater?

Tip:

Always ensure that units are consistent when performing calculations, especially when converting between milliliters and liters.