Introductory Chemistry: A Foundation
8th Edition
ISBN:9781285199030
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 114AP
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Question
![**Problem Statement:**
If 286.2 g of calcium sulfate is dissolved in enough water to make 228.3 mL of solution, what is the molarity of the solution?
(Note: Below this text, you would typically find an input box or space for students to write their answers.)
**Explanation for Students:**
*Molarity Calculation:*
1. Determine the molar mass of calcium sulfate (CaSO₄):
- Ca (Calcium): 40.08 g/mol
- S (Sulfur): 32.06 g/mol
- O₄ (Oxygen): 4 x 16.00 g/mol = 64.00 g/mol
- Total Molar Mass of CaSO₄ = 40.08 g/mol + 32.06 g/mol + 64.00 g/mol = 136.14 g/mol
2. Convert the mass of calcium sulfate to moles:
\[
\text{Moles of CaSO}_4 = \frac{286.2 \text{ g}}{136.14 \text{ g/mol}} = 2.101 \text{ moles}
\]
3. Convert the volume of solution to liters:
\[
228.3 \text{ mL} = 0.2283 \text{ L}
\]
4. Calculate the molarity of the solution:
\[
\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}} = \frac{2.101 \text{ moles}}{0.2283 \text{ L}} = 9.20 \text{ M}
\]
This is the molarity of the calcium sulfate solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d0fca7b-87e2-42be-9458-260269753889%2F14e9d414-1c86-413c-a718-b11edd279e74%2Fg9c0huo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
If 286.2 g of calcium sulfate is dissolved in enough water to make 228.3 mL of solution, what is the molarity of the solution?
(Note: Below this text, you would typically find an input box or space for students to write their answers.)
**Explanation for Students:**
*Molarity Calculation:*
1. Determine the molar mass of calcium sulfate (CaSO₄):
- Ca (Calcium): 40.08 g/mol
- S (Sulfur): 32.06 g/mol
- O₄ (Oxygen): 4 x 16.00 g/mol = 64.00 g/mol
- Total Molar Mass of CaSO₄ = 40.08 g/mol + 32.06 g/mol + 64.00 g/mol = 136.14 g/mol
2. Convert the mass of calcium sulfate to moles:
\[
\text{Moles of CaSO}_4 = \frac{286.2 \text{ g}}{136.14 \text{ g/mol}} = 2.101 \text{ moles}
\]
3. Convert the volume of solution to liters:
\[
228.3 \text{ mL} = 0.2283 \text{ L}
\]
4. Calculate the molarity of the solution:
\[
\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}} = \frac{2.101 \text{ moles}}{0.2283 \text{ L}} = 9.20 \text{ M}
\]
This is the molarity of the calcium sulfate solution.
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