5. 320g of MgCl2₂ was used to make 0.5 M solution. What volume of solution would have been made. Mol. wt. of MgCl₂ is 95. atniog llut top of how were li
5. 320g of MgCl2₂ was used to make 0.5 M solution. What volume of solution would have been made. Mol. wt. of MgCl₂ is 95. atniog llut top of how were li
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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I’ve been having difficulties with this question for homework on biology I’m not sure if I’m supposed to do the M= equation or do Grams - Mol
![### Educational Exercise: Solution Preparation Calculation
**Problem Statement:**
A quantity of 320g of MgCl₂ was used to make a 0.5 M solution. What volume of solution would have been made? *Molecular weight (Mol. wt.) of MgCl₂ is 95.*
### Steps to Solve the Problem:
1. **Calculate Moles of MgCl₂:**
To find the moles in 320g of MgCl₂, use the formula:
\[
\text{moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}}
\]
\[
\text{moles} = \frac{320g}{95g/mol} = \frac{320}{95} \approx 3.37 \text{ moles}
\]
2. **Determine the Volume of Solution:**
The molarity (M) formula is:
\[
M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}
\]
Rearrange this to find the volume:
\[
\text{volume (L)} = \frac{\text{moles of solute}}{M}
\]
\[
\text{volume (L)} = \frac{3.37 \text{ moles}}{0.5 \text{ M}} = \frac{3.37}{0.5} = 6.74 \text{ L}
\]
### Conclusion:
The volume of the solution made with 320g of MgCl₂ at a concentration of 0.5 M is **6.74 liters**.
This exercise demonstrates the application of molarity and molecular weight in practical chemistry problems, emphasizing the importance of accurate calculations in solution preparation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8a7d2ddc-7104-4036-9775-2f9910a0c3cf%2Faedcc795-f2d1-4d81-962a-3933cf273491%2F11ijozi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Exercise: Solution Preparation Calculation
**Problem Statement:**
A quantity of 320g of MgCl₂ was used to make a 0.5 M solution. What volume of solution would have been made? *Molecular weight (Mol. wt.) of MgCl₂ is 95.*
### Steps to Solve the Problem:
1. **Calculate Moles of MgCl₂:**
To find the moles in 320g of MgCl₂, use the formula:
\[
\text{moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}}
\]
\[
\text{moles} = \frac{320g}{95g/mol} = \frac{320}{95} \approx 3.37 \text{ moles}
\]
2. **Determine the Volume of Solution:**
The molarity (M) formula is:
\[
M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}
\]
Rearrange this to find the volume:
\[
\text{volume (L)} = \frac{\text{moles of solute}}{M}
\]
\[
\text{volume (L)} = \frac{3.37 \text{ moles}}{0.5 \text{ M}} = \frac{3.37}{0.5} = 6.74 \text{ L}
\]
### Conclusion:
The volume of the solution made with 320g of MgCl₂ at a concentration of 0.5 M is **6.74 liters**.
This exercise demonstrates the application of molarity and molecular weight in practical chemistry problems, emphasizing the importance of accurate calculations in solution preparation.
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