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Chapter 4: Problem 9

You need to make \(150.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{NaCl}\) solution. You have solid \(\mathrm{NaCl}\), and your lab partner has a \(2.5 \mathrm{M} \mathrm{NaCl}\) solution. Explain how you each make the \(0.10 \mathrm{M} \mathrm{NaCl}\) solution.

Short Answer

Expert verified
To make a 150.0 mL of a 0.10 M NaCl solution using solid NaCl: weigh 0.877 g of solid NaCl, add it to a 150 mL volumetric flask, fill the flask with distilled water up to the 150 mL mark, and stir to dissolve. To make it using the 2.5 M NaCl solution: measure 6 mL of the 2.5 M NaCl solution using a graduated pipette, transfer it to a 150 mL volumetric flask, fill the flask with distilled water up to the 150 mL mark, and stir to ensure even distribution.

Step by step solution

01

Calculate the required moles of NaCl using the M and V of the desired solution

To make a 0.10 M NaCl solution, we need to calculate the moles of NaCl required, using the given molarity and volume information. The formula to find moles is: Moles of solute = Molarity × Volume (in liters) Moles of NaCl = \(0.10 \mathrm{M} \times 0.150 \mathrm{L}\) Moles of NaCl = 0.015 mol Therefore, we need 0.015 moles of NaCl to make 150 mL of a 0.10 M NaCl solution.
02

Determine the amount of solid NaCl needed

To find out the amount of solid NaCl needed, we must use the molar mass of NaCl, which is 58.44 g/mol. Multiply the needed moles with the molar mass to obtain the mass of NaCl required. Mass of NaCl = Moles of NaCl × Molar Mass of NaCl Mass of NaCl = \(0.015 \mathrm{mol} \times 58.44 \frac{\mathrm{g}}{\mathrm{mol}}\) Mass of NaCl = 0.877 g Thus, 0.877 grams of solid NaCl is required for making the 0.10 M NaCl solution.
03

Find out the volume of the 2.5 M NaCl solution needed

To use the 2.5 M NaCl solution provided by the lab partner, we need to find the required volume to make a 0.10 M solution. To do this, we would use the formula: \(C_1V_1 = C_2V_2\) Where \(C_1\) and \(V_1\) represent the initial concentration and volume, and \(C_2\) and \(V_2\) represent the final concentration and volume, respectively. Rearranging the formula, we get: \(V_1 = \frac{C_2V_2}{C_1}\) Now, we will plug in the values: \(V_1 = \frac{0.1 \mathrm{M} \times 0.150 \mathrm{L}}{2.5 \mathrm{M}}\) \(V_1 = 0.006 \mathrm{L}\) \(V_1 = 6 \mathrm{mL}\) Hence, 6 mL of the 2.5 M NaCl solution is needed.
04

Making the 0.10 M NaCl solution using both methods

For Method 1, using solid NaCl: 1. Weigh 0.877 grams of solid NaCl. 2. Add the solid NaCl to a 150 mL volumetric flask. 3. Fill the flask with distilled water until the 150 mL mark is reached. 4. Stir the solution to ensure that the NaCl is completely dissolved to obtain a 0.10 M NaCl solution. For Method 2, using the 2.5 M NaCl solution: 1. Use a graduated pipette to measure 6 mL of the 2.5 M NaCl solution. 2. Transfer the 6 mL of the NaCl solution to a 150 mL volumetric flask. 3. Fill the flask with distilled water until the 150 mL mark is reached. 4. Stir the solution to ensure the NaCl solution is evenly distributed to make a 0.10 M NaCl solution.

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Most popular questions from this chapter

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50 M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 M)\) sulfuric acid b. \(\mathrm{HCl}\) from "concentrated" \((12 \mathrm{M})\) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

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A \(100.0-\mathrm{mL}\) aliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 M\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.
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