Calculate The Final Concentration When 1.35 L Of A 4.00 M Nitric Acid Solution Is Diluted To 2.75 L.${ M_1 = 4.00 , \text{M} }$ { V_1 = 1.35 \, \text{L} \} ${ M_2 = ? }$ { V_2 = 2.75 \, \text{L} \}

Understanding the Problem

In chemistry, dilution is a process where a solution of a certain concentration is mixed with a solvent to achieve a desired concentration. In this problem, we are given a 4.00 M nitric acid solution that is diluted from 1.35 L to 2.75 L. Our goal is to calculate the final concentration of the solution after dilution.

The Molarity Formula

Molarity (M) is a measure of the concentration of a solution, expressed as the number of moles of solute per liter of solution. The formula for molarity is:

${ M = \frac{\text{moles of solute}}{\text{liters of solution}} }$

In this case, we are given the initial molarity (M1) and the initial volume (V1), and we need to find the final molarity (M2) after dilution.

The Dilution Formula

The dilution formula is used to calculate the final concentration of a solution after dilution. The formula is:

${ M_2 = M_1 \times \frac{V_1}{V_2} }$

Where:

• M1 is the initial molarity
• V1 is the initial volume
• V2 is the final volume
• M2 is the final molarity

Plugging in the Values

Now that we have the formula, let's plug in the values given in the problem:

{ M_1 = 4.00 \, \text{M} \} { V_1 = 1.35 \, \text{L} \} { V_2 = 2.75 \, \text{L} \}

Substituting these values into the dilution formula, we get:

${ M_2 = 4.00 \, \text{M} \times \frac{1.35 \, \text{L}}{2.75 \, \text{L}} }$

Simplifying the Expression

To simplify the expression, we can multiply the numbers and divide the units:

${ M_2 = 4.00 \, \text{M} \times 0.4909 }$

${ M_2 = 1.9636 \, \text{M} }$

Rounding the Answer

Since we are dealing with a concentration, we can round the answer to two significant figures:

${ M_2 = 1.96 \, \text{M} }$

Conclusion

In this problem, we used the dilution formula to calculate the final concentration of a solution after dilution. We plugged in the given values, simplified the expression, and rounded the answer to two significant figures. The final concentration of the solution after dilution is 1.96 M.

Real-World Applications

Dilution is a common process in chemistry and is used in a variety of applications, including:

• Preparing solutions for laboratory experiments
• Diluting strong acids or bases to make them safer to handle
• Preparing solutions for medical use
• Preparing solutions for industrial use

Tips and Tricks

When working with dilution problems, it's essential to:

• Always check the units of the given values
• Use the correct formula for dilution
• Simplify the expression before rounding the answer
• Round the answer to the correct number of significant figures

Understanding the Problem

In chemistry, dilution is a process where a solution of a certain concentration is mixed with a solvent to achieve a desired concentration. In this problem, we are given a 4.00 M nitric acid solution that is diluted from 1.35 L to 2.75 L. Our goal is to calculate the final concentration of the solution after dilution.

The Molarity Formula

Molarity (M) is a measure of the concentration of a solution, expressed as the number of moles of solute per liter of solution. The formula for molarity is:

${ M = \frac{\text{moles of solute}}{\text{liters of solution}} }$

In this case, we are given the initial molarity (M1) and the initial volume (V1), and we need to find the final molarity (M2) after dilution.

The Dilution Formula

The dilution formula is used to calculate the final concentration of a solution after dilution. The formula is:

${ M_2 = M_1 \times \frac{V_1}{V_2} }$

Where:

• M1 is the initial molarity
• V1 is the initial volume
• V2 is the final volume
• M2 is the final molarity

Plugging in the Values

Now that we have the formula, let's plug in the values given in the problem:

{ M_1 = 4.00 \, \text{M} \} { V_1 = 1.35 \, \text{L} \} { V_2 = 2.75 \, \text{L} \}

Substituting these values into the dilution formula, we get:

${ M_2 = 4.00 \, \text{M} \times \frac{1.35 \, \text{L}}{2.75 \, \text{L}} }$

Simplifying the Expression

To simplify the expression, we can multiply the numbers and divide the units:

${ M_2 = 4.00 \, \text{M} \times 0.4909 }$

${ M_2 = 1.9636 \, \text{M} }$

Rounding the Answer

Since we are dealing with a concentration, we can round the answer to two significant figures:

${ M_2 = 1.96 \, \text{M} }$

Conclusion

In this problem, we used the dilution formula to calculate the final concentration of a solution after dilution. We plugged in the given values, simplified the expression, and rounded the answer to two significant figures. The final concentration of the solution after dilution is 1.96 M.

Real-World Applications

Dilution is a common process in chemistry and is used in a variety of applications, including:

• Preparing solutions for laboratory experiments
• Diluting strong acids or bases to make them safer to handle
• Preparing solutions for medical use
• Preparing solutions for industrial use

Tips and Tricks

When working with dilution problems, it's essential to:

• Always check the units of the given values
• Use the correct formula for dilution
• Simplify the expression before rounding the answer
• Round the answer to the correct number of significant figures

By following these tips and tricks, you can ensure that you get the correct answer and develop a deeper understanding of the concept of dilution.

Q&A: Calculating Final Concentration After Dilution

Q: What is the formula for calculating the final concentration of a solution after dilution? A: The formula for calculating the final concentration of a solution after dilution is:

${ M_2 = M_1 \times \frac{V_1}{V_2} }$

Q: What is the initial molarity (M1) in the given problem? A: The initial molarity (M1) in the given problem is 4.00 M.

Q: What is the initial volume (V1) in the given problem? A: The initial volume (V1) in the given problem is 1.35 L.

Q: What is the final volume (V2) in the given problem? A: The final volume (V2) in the given problem is 2.75 L.

Q: How do I simplify the expression in the dilution formula? A: To simplify the expression in the dilution formula, you can multiply the numbers and divide the units.

Q: How do I round the answer to the correct number of significant figures? A: To round the answer to the correct number of significant figures, you should look at the first digit after the decimal point. If it is less than 5, you round down. If it is 5 or greater, you round up.

Q: What are some real-world applications of dilution? A: Some real-world applications of dilution include preparing solutions for laboratory experiments, diluting strong acids or bases to make them safer to handle, preparing solutions for medical use, and preparing solutions for industrial use.

Q: What are some tips and tricks for working with dilution problems? A: Some tips and tricks for working with dilution problems include always checking the units of the given values, using the correct formula for dilution, simplifying the expression before rounding the answer, and rounding the answer to the correct number of significant figures.