Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Problem Statement:**
A 42.2 g sample of groundwater is found to have a lead concentration of 17.77 ppm. Calculate the mass of lead in the sample in milligrams.
**Solution:**
To find the mass of lead in the sample:
1. **Understand ppm (parts per million):**
- 1 ppm means 1 part of solute per 1 million parts of solution.
- In terms of mass, it means 1 mg of solute per 1 kg (1000 g) of solution.
2. **Calculate the mass of lead:**
- Given:
- Lead concentration = 17.77 ppm
- Sample mass = 42.2 g
- Convert the concentration to milligrams:
\[
\text{Mass of lead (mg)} = \left(\frac{\text{Concentration (ppm)}}{1000} \right) \times \text{Sample mass (g)}
\]
\[
\text{Mass of lead (mg)} = \left(\frac{17.77}{1000}\right) \times 42.2 = 0.749794 \, \text{mg}
\]
3. **Calculation Result:**
- The mass of lead in the groundwater sample is approximately 0.75 mg.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd15a2d56-c982-4a4b-8f12-312dac46f35d%2F5a271a75-d797-48ea-895c-e4754a3a59c2%2Fa7twteq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 42.2 g sample of groundwater is found to have a lead concentration of 17.77 ppm. Calculate the mass of lead in the sample in milligrams.
**Solution:**
To find the mass of lead in the sample:
1. **Understand ppm (parts per million):**
- 1 ppm means 1 part of solute per 1 million parts of solution.
- In terms of mass, it means 1 mg of solute per 1 kg (1000 g) of solution.
2. **Calculate the mass of lead:**
- Given:
- Lead concentration = 17.77 ppm
- Sample mass = 42.2 g
- Convert the concentration to milligrams:
\[
\text{Mass of lead (mg)} = \left(\frac{\text{Concentration (ppm)}}{1000} \right) \times \text{Sample mass (g)}
\]
\[
\text{Mass of lead (mg)} = \left(\frac{17.77}{1000}\right) \times 42.2 = 0.749794 \, \text{mg}
\]
3. **Calculation Result:**
- The mass of lead in the groundwater sample is approximately 0.75 mg.
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