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27. How many grams of Ba(OH)2 must be dissolved in 250 mL of solution to give a 0.0153 M solution? a. 0.00383 g e. 655 g b. 0.655 g c. 2.62 g d. 3.83 g

27. How many grams of Ba(OH)2 must be dissolved in 250 mL of solution to give a 0.0153 M solution? a. 0.00383 g e. 655 g b. 0.655 g c. 2.62 g d. 3.83 g

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
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**Question 27: Calculating the Mass of Ba(OH)₂ for Solution Preparation** **Question:** How many grams of Ba(OH)₂ must be dissolved in 250 mL of solution to give a 0.0153 M solution? **Options:** a. 0.00383 g b. 0.655 g c. 2.62 g d. 3.83 g e. 655 g To solve this problem, we need to determine the mass of barium hydroxide (Ba(OH)₂) needed to prepare 250 mL of a 0.0153 M solution. **Detailed Solution Steps:** 1. **Identify the Molarity (M)**: Given: M = 0.0153 M 2. **Convert Volume to Liters:** Given: Volume = 250 mL Convert to Liters: \( 250 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.250 \, \text{L} \) 3. **Calculate the Moles of Ba(OH)₂ Required**: Using the formula \( M = \frac{\text{moles}}{\text{Liters}} \), rearrange to find: \[ \text{Moles of Ba(OH)}_2 = M \times \text{Volume in Liters} \] \[ \text{Moles of Ba(OH)}_2 = 0.0153 \, \text{M} \times 0.250 \, \text{L} \] \[ \text{Moles of Ba(OH)}_2 = 0.003825 \, \text{moles} \] 4. **Calculate the Molar Mass of Ba(OH)₂**: - Barium (Ba): 137.33 g/mol - Oxygen (O): 16.00 g/mol (16.00 g/mol \(\times\) 2 = 32.00 g/mol) - Hydrogen (H): 1.01 g/mol (1.01 g/mol \(\times\) 2 = 2.02 g/mol) Total Molar Mass of Ba(OH)₂:
Transcribed Image Text:**Question 27: Calculating the Mass of Ba(OH)₂ for Solution Preparation** **Question:** How many grams of Ba(OH)₂ must be dissolved in 250 mL of solution to give a 0.0153 M solution? **Options:** a. 0.00383 g b. 0.655 g c. 2.62 g d. 3.83 g e. 655 g To solve this problem, we need to determine the mass of barium hydroxide (Ba(OH)₂) needed to prepare 250 mL of a 0.0153 M solution. **Detailed Solution Steps:** 1. **Identify the Molarity (M)**: Given: M = 0.0153 M 2. **Convert Volume to Liters:** Given: Volume = 250 mL Convert to Liters: \( 250 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.250 \, \text{L} \) 3. **Calculate the Moles of Ba(OH)₂ Required**: Using the formula \( M = \frac{\text{moles}}{\text{Liters}} \), rearrange to find: \[ \text{Moles of Ba(OH)}_2 = M \times \text{Volume in Liters} \] \[ \text{Moles of Ba(OH)}_2 = 0.0153 \, \text{M} \times 0.250 \, \text{L} \] \[ \text{Moles of Ba(OH)}_2 = 0.003825 \, \text{moles} \] 4. **Calculate the Molar Mass of Ba(OH)₂**: - Barium (Ba): 137.33 g/mol - Oxygen (O): 16.00 g/mol (16.00 g/mol \(\times\) 2 = 32.00 g/mol) - Hydrogen (H): 1.01 g/mol (1.01 g/mol \(\times\) 2 = 2.02 g/mol) Total Molar Mass of Ba(OH)₂:
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