Suppose 1.77 g of barium nitrate is dissolved in 350. ml of a s2.0 m M aqueous solution of sodium chromate. Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. Round your answer to 3 significant digits.
Suppose 1.77 g of barium nitrate is dissolved in 350. ml of a s2.0 m M aqueous solution of sodium chromate. Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. Round your answer to 3 significant digits.
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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![### Problem Statement
**Suppose 1.77 grams of barium nitrate is dissolved in 350 milliliters of a 5.20 millimolar aqueous solution of sodium chromate.**
**Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it.**
**Round your answer to 3 significant digits.**
---
### Solution Box
\[ \text{M} \]
---
### Explanation
This problem asks you to calculate the final molarity of the barium cation in a mixed solution. Here are the steps to solve the problem:
1. **Convert grams of barium nitrate to moles**:
\[
\text{Molecular weight of Ba(NO}_3\text{)}_2 = 137.33 \, (\text{Ba}) + 2[14.01 \, (\text{N}) + 3 \times 16.00 \, (\text{O})] = 261.34 \, \text{g/mol}
\]
\[
\text{Moles of Ba(NO}_3\text{)}_2 = \frac{1.77 \, \text{g}}{261.34 \, \text{g/mol}} = 0.00677 \, \text{mol}
\]
2. **Determine the final volume in liters**:
\[
350 \, \text{mL} = 0.350 \, \text{L}
\]
3. **Calculate the final molarity (M)**:
\[
\text{Molarity of Ba}^{2+} = \frac{\text{Moles of Ba}^{2+}}{\text{Volume of solution in L}} = \frac{0.00677 \, \text{mol}}{0.350 \, \text{L}} = 0.0193 \, \text{M}
\]
### Final Answer
\(\boxed{0.0193 \, \text{M}}\)
---
*Note: There were no graphs or diagrams to describe in detail for this problem.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fada93b44-8b7d-4f67-9ba6-5f5a01e75f14%2Fb4714305-02b0-43ac-a13f-0be7afe22fd6%2Fyibtjyo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Suppose 1.77 grams of barium nitrate is dissolved in 350 milliliters of a 5.20 millimolar aqueous solution of sodium chromate.**
**Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it.**
**Round your answer to 3 significant digits.**
---
### Solution Box
\[ \text{M} \]
---
### Explanation
This problem asks you to calculate the final molarity of the barium cation in a mixed solution. Here are the steps to solve the problem:
1. **Convert grams of barium nitrate to moles**:
\[
\text{Molecular weight of Ba(NO}_3\text{)}_2 = 137.33 \, (\text{Ba}) + 2[14.01 \, (\text{N}) + 3 \times 16.00 \, (\text{O})] = 261.34 \, \text{g/mol}
\]
\[
\text{Moles of Ba(NO}_3\text{)}_2 = \frac{1.77 \, \text{g}}{261.34 \, \text{g/mol}} = 0.00677 \, \text{mol}
\]
2. **Determine the final volume in liters**:
\[
350 \, \text{mL} = 0.350 \, \text{L}
\]
3. **Calculate the final molarity (M)**:
\[
\text{Molarity of Ba}^{2+} = \frac{\text{Moles of Ba}^{2+}}{\text{Volume of solution in L}} = \frac{0.00677 \, \text{mol}}{0.350 \, \text{L}} = 0.0193 \, \text{M}
\]
### Final Answer
\(\boxed{0.0193 \, \text{M}}\)
---
*Note: There were no graphs or diagrams to describe in detail for this problem.*
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