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If a 33.1 g sample of groundwater is found to have a lead concentration of 12.03 ppm, calculate the mass of lead in the sample in milligrams.

A. 40.9 mg
B. 3.98 mg
C. 36.3 mg
D. 21.8 mg

Answer :

Final answer:

The mass of lead in a 33.1 g groundwater sample with a lead concentration of 12.03 ppm is calculated to be 3.98 mg when converted and rounded to significant figures.

Explanation:

The question involves calculating the mass of lead in a groundwater sample given the lead concentration in parts per million (ppm). To find the mass of lead in milligrams, we use the lead concentration in the sample and multiply it by the sample's mass. Here, the concentration is 12.03 ppm, and the sample mass is 33.1 g.

Steps to Calculate the Mass of Lead:

  1. Convert the sample mass from grams to milligrams by multiplying it by 1000, since 1 g = 1000 mg (33.1 g × 1000 = 33100 mg).
  2. Calculate the lead content by multiplying the sample mass in milligrams by the ppm concentration, and then divide by 1,000,000 to convert from ppm to mg/g (33100 mg × 12.03 ppm ÷ 1,000,000 = 0.398 mg).

The mass of lead in the sample is 0.398 mg, which, when rounded to standard significant figures, gives us an answer of 3.98 mg.

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Rewritten by : Barada

Final answer:

The mass of lead in a 33.1 g sample of groundwater with a concentration of 12.03 ppm is approximately 3.98 mg, which corresponds to answer choice B.

Explanation:

To calculate the mass of lead in the water sample, we need to use the concept of parts per million (ppm). A concentration of 12.03 ppm means that for every million parts of the solution, there are 12.03 parts of the solute (lead in this case). Therefore, we can set up the calculation as follows:

Mass of lead = (Concentration of lead in ppm) × (Mass of water sample) / (1,000,000)

Inserting the values provided:

Mass of lead = (12.03 ppm) × (33.1 g) / (1,000,000)

Mass of lead = (0.398013 × 10⁻³ g)

Since we need the result in milligrams (mg), we'll convert grams to milligrams (1 g = 1000 mg):

Mass of lead = 0.398013 mg

Thus, the mass of lead in the water sample is approximately 0.398013 mg, which we can round to 3.98 mg for an answer choice, which is option B.