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In 1993 the Minnesota Department of Health set a health risk limit for chloroform in groundwater of 60.0 g/L Suppose an analytical chemist receives a sample of groundwater with a measured volume of 79.0 mL. Calculate the maximum mass in milligrams of chloroform which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits. mg

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Answer:

4.74 × 10³ mg

Explanation:

Given data

  • Health risk limit for chloroform in groundwater: 60.0 g/L
  • Volume of the sample of groundwater: 79.0 mL = 79.0 × 10⁻³ L

The maximum mass of chloroform that there could be in the sample of groundwater to meet the standards are:

79.0 × 10⁻³ L × 60.0 g/L = 4.74 g

1 gram is equal to 10³ milligrams. Then,

4.74 g × (10³ mg/1 g) = 4.74 × 10³ mg

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