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How many grams of calcium chloride are needed to make 382 mL of a 2.54 M solution?

How many grams of calcium chloride are needed to make 382 mL of a 2.54 M solution?

Chemistry
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:** How many grams of calcium chloride are needed to make 382 mL of a 2.54 M solution? **Explanation:** To determine the amount of calcium chloride (CaCl₂) needed to prepare a 2.54 M solution with a volume of 382 mL, you can follow these steps: 1. **Convert the volume from mL to L:** \[ \text{Volume (L)} = \frac{382 \text{ mL}}{1000} = 0.382 \text{ L} \] 2. **Calculate the moles of CaCl₂ needed:** \[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} = 2.54 \text{ M} \times 0.382 \text{ L} = 0.97148 \text{ moles} \] 3. **Find the molar mass of CaCl₂:** - Calcium (Ca) = 40.08 g/mol - Chlorine (Cl) = 35.45 g/mol - Total molar mass of CaCl₂ = 40.08 g/mol + (2 \times 35.45 g/mol) = 110.98 g/mol 4. **Calculate the mass of CaCl₂ needed:** \[ \text{Mass (g)} = \text{Moles} \times \text{Molar mass} = 0.97148 \text{ moles} \times 110.98 \text{ g/mol} \approx 107.79 \text{ g} \] Thus, 107.79 grams of calcium chloride are needed to make 382 mL of a 2.54 M solution.
Transcribed Image Text:**Question:** How many grams of calcium chloride are needed to make 382 mL of a 2.54 M solution? **Explanation:** To determine the amount of calcium chloride (CaCl₂) needed to prepare a 2.54 M solution with a volume of 382 mL, you can follow these steps: 1. **Convert the volume from mL to L:** \[ \text{Volume (L)} = \frac{382 \text{ mL}}{1000} = 0.382 \text{ L} \] 2. **Calculate the moles of CaCl₂ needed:** \[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} = 2.54 \text{ M} \times 0.382 \text{ L} = 0.97148 \text{ moles} \] 3. **Find the molar mass of CaCl₂:** - Calcium (Ca) = 40.08 g/mol - Chlorine (Cl) = 35.45 g/mol - Total molar mass of CaCl₂ = 40.08 g/mol + (2 \times 35.45 g/mol) = 110.98 g/mol 4. **Calculate the mass of CaCl₂ needed:** \[ \text{Mass (g)} = \text{Moles} \times \text{Molar mass} = 0.97148 \text{ moles} \times 110.98 \text{ g/mol} \approx 107.79 \text{ g} \] Thus, 107.79 grams of calcium chloride are needed to make 382 mL of a 2.54 M solution.
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