You need to make \(150.0 \mathrm{~mL}\) of a \(0.10-M \mathrm{NaCl}\) 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 \(0.10-M \mathrm{NaCl}\) solution.

Short Answer

Expert verified
To make a \(0.10-M \mathrm{NaCl}\) solution using solid \(\mathrm{NaCl}\), dissolve \(0.877 \, \mathrm{g}\) of solid \(\mathrm{NaCl}\) in \(150 \mathrm{~mL}\) of distilled water. To make the same solution using the given \(2.5-M \mathrm{NaCl}\) solution, dilute \(6.0 \mathrm{~mL}\) of the \(2.5-M \mathrm{NaCl}\) solution with distilled water in a \(150 \mathrm{~mL}\) volumetric flask.

Step by step solution

01

Preparing the \(0.10-M \mathrm{NaCl}\) solution using solid \(\mathrm{NaCl}\).

1. Calculate the moles of \(\mathrm{NaCl}\) needed for the desired concentration and volume: For a \(0.10-M \mathrm{NaCl}\) solution, the formula is: Moles of \(\mathrm{NaCl}\) = volume of solution (L) × molarity (\(M\)) First, convert \(150 \mathrm{~mL}\) to liters: \(150 \mathrm{~mL} = 0.15 \mathrm{~L}\). Moles of \(\mathrm{NaCl}\) = \(0.15 \mathrm{~L} \times 0.10-M = 0.015 \, \mathrm{mol}\). 2. Calculate the mass of solid \(\mathrm{NaCl}\) needed: Using the molar mass of \(\mathrm{NaCl}\): \(58.44 \, \mathrm{g/mol}\), Mass of \(\mathrm{NaCl}\) = moles of \(\mathrm{NaCl} \times\) molar mass Mass of \(\mathrm{NaCl} = 0.015 \, \mathrm{mol} \times 58.44 \,\mathrm{g/mol} = 0.877 \, \mathrm{g}\). 3. Preparing the solution: Weigh out \(0.877 \, \mathrm{g}\) of solid \(\mathrm{NaCl}\), and dissolve it in approximately \(100 \mathrm{~mL}\) of distilled water in a volumetric flask. Fill the volumetric flask to the \(150 \mathrm{~mL}\) mark with distilled water, and mix thoroughly to obtain the desired \(0.10-M \mathrm{NaCl}\) solution.
02

Preparing the \(0.10-M \mathrm{NaCl}\) solution using the given \(2.5-M \mathrm{NaCl}\) solution.

1. Calculate the volume of \(2.5-M \mathrm{NaCl}\) solution needed: Using dilution formula: \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the initial concentration and volume of the given solution, and \(C_2\) and \(V_2\) are the final concentration and volume of the desired solution. Rearrange the formula for \(V_1\): \(V_1 = \frac{C_2V_2}{C_1}\). 2. Calculate the required volume: Plug in the desired concentration \(C_2 = 0.10 \mathrm{~M}\), the desired volume \(V_2 = 0.15 \mathrm{~L}\), and the given concentration \(C_1 = 2.5 \mathrm{~M}\): \(V_1 = \frac{0.10-M \times 0.15 \mathrm{~L}}{2.5 \mathrm{~M}} = 0.006 \mathrm{~L} = 6.0 \mathrm{~mL}\). 3. Preparing the solution: Using a volumetric pipette, measure out \(6.0 \mathrm{~mL}\) of the \(2.5-M \mathrm{NaCl}\) solution, and transfer it to a \(150 \mathrm{~mL}\) volumetric flask. Fill the volumetric flask to the \(150 \mathrm{~mL}\) mark with distilled water, and mix thoroughly to obtain the desired \(0.10-M \mathrm{NaCl}\) solution.

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Most popular questions from this chapter

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{~S}(g)\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

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