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
![**Educational Content - Chemistry Problem**
---
**Question:**
How much calcium chloride is needed to make 1.00 L of a 0.1 M solution of calcium chloride?
**Visible Calculation:**
0.030
---
To answer this question, you need to use the concept of molarity and the molar mass of calcium chloride (CaCl₂). The molarity (M) is the number of moles of solute per liter of solution. For calcium chloride:
1. Calculate the number of moles needed:
\[
Moles = Molarity (M) \times Volume (L)
\]
\[
Moles = 0.1 \, \text{M} \times 1.00 \, \text{L}
\]
\[
Moles = 0.1 \, \text{moles}
\]
2. Calculate the mass of calcium chloride required using its molar mass:
- The molar mass of CaCl₂ = 40.08 (Ca) + 2 \times 35.453 (Cl) = 110.98 g/mol
\[
Mass = Moles \times Molar Mass
\]
\[
Mass = 0.1 \, \text{moles} \times 110.98 \, \text{g/mol}
\]
\[
Mass = 11.098 \, \text{g}
\]
Therefore, you need 11.098 grams of calcium chloride to make 1.00 L of a 0.1 M solution.
**Note:**
The number "0.030" shown in the image appears incorrect based on the standard chemical calculations for calcium chloride.
---
This section includes an image showing the handwritten problem and the number "0.030." It is important to cross-check your calculations to ensure accuracy in solutions and to understand the underlying concepts thoroughly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf4fd61a-46d9-45b2-8502-88fe4eac45cd%2F30d631d5-bbca-4aba-8354-2838cd3138f3%2Fuagmt0h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content - Chemistry Problem**
---
**Question:**
How much calcium chloride is needed to make 1.00 L of a 0.1 M solution of calcium chloride?
**Visible Calculation:**
0.030
---
To answer this question, you need to use the concept of molarity and the molar mass of calcium chloride (CaCl₂). The molarity (M) is the number of moles of solute per liter of solution. For calcium chloride:
1. Calculate the number of moles needed:
\[
Moles = Molarity (M) \times Volume (L)
\]
\[
Moles = 0.1 \, \text{M} \times 1.00 \, \text{L}
\]
\[
Moles = 0.1 \, \text{moles}
\]
2. Calculate the mass of calcium chloride required using its molar mass:
- The molar mass of CaCl₂ = 40.08 (Ca) + 2 \times 35.453 (Cl) = 110.98 g/mol
\[
Mass = Moles \times Molar Mass
\]
\[
Mass = 0.1 \, \text{moles} \times 110.98 \, \text{g/mol}
\]
\[
Mass = 11.098 \, \text{g}
\]
Therefore, you need 11.098 grams of calcium chloride to make 1.00 L of a 0.1 M solution.
**Note:**
The number "0.030" shown in the image appears incorrect based on the standard chemical calculations for calcium chloride.
---
This section includes an image showing the handwritten problem and the number "0.030." It is important to cross-check your calculations to ensure accuracy in solutions and to understand the underlying concepts thoroughly.
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