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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 can be calculated using the given information in four steps. First, find the mass of glacial acetic acid: \(Mass = 1.049 \, \frac{g}{mL} \times 20.00 \, mL\). Second, calculate the moles of acetic acid using its molar mass: Moles of acetic acid = \( \frac{Mass}{60.05 \, g/mol} \). Then, convert the total volume of the solution to liters: Total volume = \( 250.0 \, mL \times \frac{1 \, L}{1000 \, mL} \). Finally, calculate the molarity of the acetic acid solution: Molarity = \( \frac{Moles \, of \, acetic \, acid}{Total \, volume \, (L)} \).

Step by step solution

01

Find the mass of glacial acetic acid

Using the given density and volume, we can calculate the mass of the glacial acetic acid: Mass = Density 脳 Volume \(Mass = 1.049 \, \frac{g}{mL} \times 20.00 \, mL\)
02

Calculate the moles of acetic acid

Using the molar mass of acetic acid (60.05 g/mol), convert the mass of acetic acid to moles: Moles of acetic acid = \( \frac{Mass}{Molar \, mass} \) Moles of acetic acid = \( \frac{Mass}{60.05 \, g/mol} \)
03

Calculate the total volume of the solution in liters

We know that we have 20.0 mL of glacial acetic acid and enough water is added to make 250.0 mL of solution, so we need to convert the total volume from milliliters to liters: Total volume = \(250.0 \, mL \times \frac{1\, L}{1000 \, mL} \)
04

Calculate the molarity of the acetic acid solution

The molarity is defined as the number of moles of solute (acetic acid) per liter of solution. To find this, we will divide the moles of acetic acid by the total volume of the solution in liters: Molarity = \( \frac{Moles \, of\, acetic\, acid}{Total\, volume\, (L)} \)

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Most popular questions from this chapter

Specify what ions are present in solution upon dissolving each of the following substances in water: a) \(\mathrm{ZnCl}_{2}\), (b) \(\mathrm{HNO}_{3}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\), (d) \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-} ; \mathrm{CO}_{3}^{2-} ; \mathrm{NO}_{3}^{-} ?\)

When methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is dissolved in water, a nonconducting solution results. When acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), dissolves in water, the solution is weakly conducting and acidic in nature. Describe what happens upon dissolution in the two cases, and account for the different results.

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the observations made above. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or acid-base reactions. (a) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow\) $$ \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) (c) \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) $$ \mathrm{SrSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q) $$ (d) \(4 \mathrm{Zn}(s)+10 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow\) $$ 4 \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2} \mathrm{O}(g)+5 \mathrm{H}_{2} \mathrm{O}(l) $$
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