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What volume (in mL) of 0.0887 M MgF₂ solution is needed to make 275.0 mL of 0.0224 M MgF₂ solution? O 72.3 mL 10.9 mL 69.4 mL O 14.4 mL 91.8 mL

What volume (in mL) of 0.0887 M MgF₂ solution is needed to make 275.0 mL of 0.0224 M MgF₂ solution? O 72.3 mL 10.9 mL 69.4 mL O 14.4 mL 91.8 mL

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:** What volume (in mL) of 0.0887 M MgF₂ solution is needed to make 275.0 mL of 0.0224 M MgF₂ solution? **Answer Choices:** - A. 72.3 mL - B. 10.9 mL - C. 69.4 mL - D. 14.4 mL - E. 91.8 mL **Explanation:** This problem involves using the dilution equation: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the concentration of the initial solution (0.0887 M) - \( V_1 \) is the volume of the initial solution - \( C_2 \) is the concentration of the final solution (0.0224 M) - \( V_2 \) is the volume of the final solution (275.0 mL) Solve for \( V_1 \) to find the required volume of the initial solution.
Transcribed Image Text:**Question:** What volume (in mL) of 0.0887 M MgF₂ solution is needed to make 275.0 mL of 0.0224 M MgF₂ solution? **Answer Choices:** - A. 72.3 mL - B. 10.9 mL - C. 69.4 mL - D. 14.4 mL - E. 91.8 mL **Explanation:** This problem involves using the dilution equation: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the concentration of the initial solution (0.0887 M) - \( V_1 \) is the volume of the initial solution - \( C_2 \) is the concentration of the final solution (0.0224 M) - \( V_2 \) is the volume of the final solution (275.0 mL) Solve for \( V_1 \) to find the required volume of the initial solution.
**Question:** How many grams of NaCl are required to make 250.00 mL of a 3.000 M solution? **Options:** - ○ 14.60 g - ○ 175.3 g - ○ 58.40 g - ○ 43.83 g **Explanation:** To find the amount of grams needed, use the formula: \[ \text{grams of NaCl} = \text{molarity} \times \text{volume in liters} \times \text{molar mass of NaCl} \] - **Molarity (M)** = 3.000 M - **Volume (L)** = 250.00 mL = 0.250 L - **Molar mass of NaCl** ≈ 58.44 g/mol \[ \text{grams of NaCl} = 3.000 \, \text{M} \times 0.250 \, \text{L} \times 58.44 \, \text{g/mol} \] \[ \text{grams of NaCl} = 43.83 \, \text{g} \] Thus, the correct answer is 43.83 g.
Transcribed Image Text:**Question:** How many grams of NaCl are required to make 250.00 mL of a 3.000 M solution? **Options:** - ○ 14.60 g - ○ 175.3 g - ○ 58.40 g - ○ 43.83 g **Explanation:** To find the amount of grams needed, use the formula: \[ \text{grams of NaCl} = \text{molarity} \times \text{volume in liters} \times \text{molar mass of NaCl} \] - **Molarity (M)** = 3.000 M - **Volume (L)** = 250.00 mL = 0.250 L - **Molar mass of NaCl** ≈ 58.44 g/mol \[ \text{grams of NaCl} = 3.000 \, \text{M} \times 0.250 \, \text{L} \times 58.44 \, \text{g/mol} \] \[ \text{grams of NaCl} = 43.83 \, \text{g} \] Thus, the correct answer is 43.83 g.
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