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Chapter 4: Problem 77

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
The molarity of the acetic acid solution is \(1.4 \, M\).

Step by step solution

01

Determine the mass of glacial acetic acid used

To calculate the mass of acetic acid used, we will multiply the density of acetic acid with the volume of glacial acetic acid we are using for the solution: Mass = Density × Volume The density of glacial acetic acid is \(1.049 \, \frac{g}{mL}\), and the volume is \(20.00 \, mL\). Thus, Mass = \(1.049 \, \frac{g}{mL} \times 20.00 \, mL\)
02

Calculate the mass of acetic acid used

Plug in the given values to compute the mass: Mass = \(1.049 \, \frac{g}{mL} \times 20.00 \, mL = 21.0 \, g\) So, we have \(21.0 \, g\) of glacial acetic acid that will be dissolved in the solution.
03

Determine the moles of acetic acid in the solution

Now, we will find the number of moles of acetic acid dissolved in the solution. The molar mass of acetic acid is \(60.05 \, \frac{g}{mol}\). To find the moles of acetic acid, we will use the formula: Moles = \(\frac{Mass}{Molar \, Mass}\), where Mass = \(21.0 \, g\) and Molar Mass = \(60.05 \, \frac{g}{mol}\) Moles = \(\frac{21.0 \, g}{60.05 \, \frac{g}{mol}}\)
04

Calculate the moles of acetic acid in the solution

Plug the given values of acetic acid to get the moles: Moles = \(\frac{21.0 \, g}{60.05 \, \frac{g}{mol}} = 0.35 \, mol\) So there are \(0.35 \, mol\) of acetic acid in the solution.
05

Calculate the molarity of the solution

Finally, we'll find the molarity of the solution using the formula: Molarity = \(\frac{Moles}{Volume \, of \, solution}\), where Moles = \(0.35 \, mol\) and the Volume of solution = \(250.0 \, mL\) We need to convert mL to L to keep the units consistent: \(250.0 \, mL = 0.250 \, L\) Molarity = \(\frac{0.35 \, mol}{0.250 \, L}\)
06

Find the molarity of the solution

Divide the given values to compute the molarity of the solution: Molarity = \(\frac{0.35 \, mol}{0.250 \, L} = 1.4 \, M\) The molarity of the acetic acid solution is \(1.4 \, M\).

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Most popular questions from this chapter

Write balanced molecular and net ionic equations for the reactions of (a) hydrochloric acid with nickel, (b) dilute sulfuric acid with iron, \((\mathbf{c})\) hydrobromic acid with magnesium, (d) acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) with zinc.

Calicheamicin gamma-1, $\mathrm{C}_{\mathrm{ss}} \mathrm{H}_{74} \mathrm{IN}_{3} \mathrm{O}_{21} \mathrm{~S}_{4},$ is one of the most potent antibiotics known: one molecule kills one bacterial cell. Describe how you would (carefully!) prepare \(25.00 \mathrm{~mL}\) of an aqueous calicheamicin gamma- 1 solution that could kill \(1.0 \times 10^{8}\) bacteria, starting from a \(5.00 \times 10^{-9} \mathrm{M}\) stock solution of the antibiotic.

A sample of \(8.69 \mathrm{~g}\) of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(155.0 \mathrm{~mL}\) of \(0.750 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .\) (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Zn}(\mathrm{OH})_{2}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{ZnSO}_{4}\) are present after the reaction is complete?

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could be present in the unknown salt solution: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+} ?\)

Because the oxide ion is basic, metal oxides react readily with acids. (a) Write the net ionic equation for the following reaction: $$ \mathrm{FeO}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (b) Based on the equation in part (a), write the net ionic equation for the reaction that occurs between \(\mathrm{NiO}(s)\) and an aqueous solution of nitric acid.
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