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 4:**
A student needs to prepare 210.0 mL of a 1.99 molar solution (1.99 M) of potassium chloride. How many grams of KCl will be needed?
**Explanation:**
In this problem, you need to calculate the mass of potassium chloride (KCl) required to make a specific solution. The necessary steps involve using the molarity equation and converting the volume from milliliters to liters.
1. **Convert volume from mL to L:**
\( 210.0 \, \text{mL} = 0.210 \, \text{L} \)
2. **Use the molarity formula:**
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Rearrange to solve for moles of solute:
\[ \text{moles of KCl} = \text{Molarity} \times \text{Liters of Solution} \]
3. **Calculate moles of KCl:**
Moles of KCl = \( 1.99 \, \text{M} \times 0.210 \, \text{L} \)
Moles of KCl = \( 0.4179 \, \text{moles} \)
4. **Calculate grams of KCl:**
\[ \text{Mass (g)} = \text{moles} \times \text{molar mass of KCl} \]
Using the molar mass of KCl (approximately 74.55 g/mol):
\[ \text{Mass} = 0.4179 \, \text{moles} \times 74.55 \, \text{g/mol} \]
\[ \text{Mass} \approx 31.14 \, \text{grams of KCl} \]
The student will need approximately 31.14 grams of KCl to prepare the solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabd48a79-2397-457e-8505-40fa8e1d092f%2F6f10a977-d1a0-4718-beb4-82a84c5cb802%2Flryz9m3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 4:**
A student needs to prepare 210.0 mL of a 1.99 molar solution (1.99 M) of potassium chloride. How many grams of KCl will be needed?
**Explanation:**
In this problem, you need to calculate the mass of potassium chloride (KCl) required to make a specific solution. The necessary steps involve using the molarity equation and converting the volume from milliliters to liters.
1. **Convert volume from mL to L:**
\( 210.0 \, \text{mL} = 0.210 \, \text{L} \)
2. **Use the molarity formula:**
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Rearrange to solve for moles of solute:
\[ \text{moles of KCl} = \text{Molarity} \times \text{Liters of Solution} \]
3. **Calculate moles of KCl:**
Moles of KCl = \( 1.99 \, \text{M} \times 0.210 \, \text{L} \)
Moles of KCl = \( 0.4179 \, \text{moles} \)
4. **Calculate grams of KCl:**
\[ \text{Mass (g)} = \text{moles} \times \text{molar mass of KCl} \]
Using the molar mass of KCl (approximately 74.55 g/mol):
\[ \text{Mass} = 0.4179 \, \text{moles} \times 74.55 \, \text{g/mol} \]
\[ \text{Mass} \approx 31.14 \, \text{grams of KCl} \]
The student will need approximately 31.14 grams of KCl to prepare the solution.
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