**Question:**If 17.9 g of KBr (Molar Mass = 119.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the concentration of KBr in the resulting solution?---**Explanation:**To determine the concentration of KBr in the solution, we use the formula for molarity (M), which is the number of moles of solute per liter of solution:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]1. **Calculate moles of KBr:** - Given: 17.9 g of KBr - Molar Mass of KBr = 119.00 g/mol \[ \text{Moles of KBr} = \frac{17.9 \text{ g}}{119.00 \text{ g/mol}} = 0.1504 \text{ mol} \]2. **Convert the volume of solution to liters:** - Volume = 500.0 mL = 0.500 L3. **Calculate molarity:** \[ \text{Molarity (M)} = \frac{0.1504 \text{ mol}}{0.500 \text{ L}} = 0.3008 \text{ mol/L} \]**Conclusion:**The concentration of KBr in the resulting solution is 0.3008 M.---**Additional Resources:**For further reference on concentration calculations, consider exploring online chemistry tutorials or educational modules available on this platform.

Molarity M : It is the number of moles of solute dissolved per litre of the…

Answered: If 17.9 g of KBr (MM = 119.00 g/mol)…

If 17.9 g of KBr (MM = 119.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the concentration of KBr in the resulting solution?

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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Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

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