---### Calculating the Volume of a Silver Nitrate Solution**Problem Statement:**If 20. g of \( \text{AgNO}_3 \) is available, what volume of 0.20 M \( \text{AgNO}_3 \) solution can be prepared?**Solution:**---#### Given:- **Mass of \( \text{AgNO}_3 \)** = 20 g- **Molarity (M)** of the solution = 0.20 M#### Required:- **Volume of the solution** in mL **Formula to use:**\[ \text{Volume (L)} = \frac{\text{Mass of } \text{AgNO}_3 \text{ (g)}}{\text{Molarity (M)} \times \text{Molar Mass of } \text{AgNO}_3 \text{ (g/mol)}} \]**Molar Mass of \( \text{AgNO}_3 \):**- Silver (Ag): 107.87 g/mol- Nitrogen (N): 14.01 g/mol- Oxygen (O): 16.00 g/mol (since there are 3 oxygen atoms, it's 3 x 16.00)So,\[ \text{Molar Mass of } \text{AgNO}_3 = 107.87 + 14.01 + (3 \times 16) \ g/mol \]**Calculate the molar mass:**\[ \text{Molar Mass of } \text{AgNO}_3 = 107.87 + 14.01 + 48.00 \ g/mol \]\[ \text{Molar Mass of } \text{AgNO}_3 = 169.88 \ g/mol \]**Calculate the volume:**\[ \text{Volume (L)} = \frac{20 \ \text{g}}{0.20 \ \text{M} \times 169.88 \ \text{g/mol}} \]\[ \text{Volume (L)} = \frac{20}{33.976} \]\[ \text{Volume (L)} \approx 0.588 \]To convert this volume to mL:\[ \text{Volume (mL)} = 0.588 \ \text{L} \times 1000 \ \text{mL/L} \] **Question:**A solution was prepared by mixing \( 40.00 \) mL of \( 0.100 \, M \) \( \text{HNO}_3 \) and \( 170.00 \) mL of \( 0.200 \, M \) \( \text{HNO}_3 \). Calculate the molarity of the final solution of nitric acid.Molarity = ____ \( M \)**Explanation:**To calculate the final molarity of the nitric acid solution, you need to use the formula for combined solutions:\[ C_{\text{final}} = \frac{(C_1 \times V_1) + (C_2 \times V_2)}{V_{\text{total}}} \]Where:- \( C_1 \) and \( C_2 \) are the initial concentrations of the two solutions.- \( V_1 \) and \( V_2 \) are the volumes of the two solutions.- \( V_{\text{total}} \) is the total volume of the final solution.In this example:- \( C_1 = 0.100 \, M \)- \( V_1 = 40.00 \, mL \)- \( C_2 = 0.200 \, M \)- \( V_2 = 170.00 \, mL \)Calculate the moles of \( \text{HNO}_3 \) in each solution:- Moles in first solution \( = C_1 \times V_1 = 0.100 \, M \times 40.00 \times 10^{-3} \, L = 0.00400 \, \text{moles} \)- Moles in second solution \( = C_2 \times V_2 = 0.200 \, M \times 170.00 \times 10^{-3} \, L = 0.03400 \, \text{moles} \)Total moles of \( \text{HNO}_3 = 0.00400 + 0.03400 = 0.03800 \, \text{moles} \)Total volume \( V_{\text{total}} = V_1 + V_2 = 40.00 \, mL + 170.00 \, mL =

To solve this problem we have to calculate the volume and molarity of the solution

Answered: If 20. g of AgNO3 is available, what…

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If 20. g of AgNO3 is available, what volume of 0.20 M AgNO3 solution can be prepared? Volume = mL

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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