A chemist wants to make 5.5 L of a 0.300 M CaCl2 solution. What mass of CaCl2 (in g) should the chemist use?
Ch.5 - Introduction to Solutions and Aqueous Solutions
Chapter 5, Problem 31
To what volume should you dilute 50.0 mL of a 12 M stock HNO3 solution to obtain a 0.100 M HNO3 solution?
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Identify the dilution formula: \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( V_1 \) are the concentration and volume of the stock solution, and \( C_2 \) and \( V_2 \) are the concentration and volume of the diluted solution.
Substitute the known values into the formula: \( 12 \text{ M} \times 50.0 \text{ mL} = 0.100 \text{ M} \times V_2 \).
Rearrange the equation to solve for \( V_2 \): \( V_2 = \frac{12 \text{ M} \times 50.0 \text{ mL}}{0.100 \text{ M}} \).
Calculate the value of \( V_2 \) to find the total volume of the diluted solution.
The result will give you the volume to which you need to dilute the stock solution to achieve the desired concentration.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Dilution Principle
The dilution principle states that when a solution is diluted, the number of moles of solute remains constant. This can be expressed mathematically as C1V1 = C2V2, where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the diluted solution.
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Molarity (M)
Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L). Understanding molarity is crucial for calculating how much of a stock solution is needed to achieve a desired concentration in a diluted solution.
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Volume Calculation
Volume calculation involves determining the final volume of a solution after dilution. In the context of the dilution principle, once the initial volume and concentration are known, the final volume can be calculated using the rearranged formula V2 = (C1V1) / C2, allowing for the preparation of solutions with specific concentrations.
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