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What volume of \(2.06 \mathrm{M} \mathrm{KMnO}_{4},\) in liters, contains \(322 \mathrm{g}\) of solute?

Short Answer

Expert verified
0.989 liters of 2.06 M KMnO4 is needed.

Step by step solution

01

Calculate Molar Mass of KMnO4

Firstly, we need to find the molar mass of potassium permanganate (\( ext{KMnO}_4\)). Calculate it by adding the atomic masses of all elements: \(\text{K} = 39.10\ \text{g/mol}, \text{Mn} = 54.94\ \text{g/mol}, \text{O} = 16.00\ \text{g/mol}\). The molar mass of \(\text{KMnO}_4\) is thus \(39.10 + 54.94 + 4 \times 16.00 = 158.04\ \text{g/mol}\).
02

Calculate Moles of Solute

Next, use the molar mass to find the number of moles of \(\text{KMnO}_4\) in \(322\ \text{g}\). Use the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). So, \(\text{moles} = \frac{322}{158.04} = 2.037\, \text{mol}\).
03

Calculate Volume of Solution Needed

Now that we know the moles of solute, we can use the molarity formula to find the volume of \(\text{KMnO}_4\). Molarity is given by the formula \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\). Rearranging gives \(\text{volume} = \frac{\text{moles of solute}}{\text{Molarity}}\). So, \(\text{volume} = \frac{2.037}{2.06} = 0.989\, \text{L}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Calculating molar mass is essential in solution chemistry because it allows us to determine how much of a compound is needed to achieve a desired concentration in a solution. The molar mass is the sum of the atomic masses of all atoms in a molecule. For example, to compute the molar mass of potassium permanganate (KMnO鈧), we add together the molar masses of potassium (K), manganese (Mn), and oxygen (O):
- Potassium (K) contributes 39.10 g/mol.
- Manganese (Mn) contributes 54.94 g/mol.
- Each Oxygen (O) contributes 16.00 g/mol, and since there are four oxygen atoms, we multiply 16.00 by 4 to get 64.00 g/mol.
Adding these together, the molar mass of KMnO鈧 is 158.04 g/mol. Once calculated, this value is used to relate the mass of a compound to the number of moles, which is a measure of the quantity of substance.
Potassium Permanganate
Potassium permanganate, represented as KMnO鈧, is a chemical compound composed of potassium, manganese, and oxygen. Known for its striking deep purple color, KMnO鈧 is often used in chemistry as an oxidizing agent. It's highly soluble in water and dissociates into K鈦 and MnO鈧勨伝 ions when dissolved.
KMnO鈧 has a variety of applications beyond the laboratory:
  • It serves as a disinfectant and antiseptic in medical and sanitation practices.
  • In aquariums, it's used to treat certain types of fish diseases.
  • In water treatment facilities, it's utilized to treat iron and hydrogen sulfide.
Given its oxidizing properties, it's important to handle KMnO鈧 with care, as it can stain skin and surfaces and may cause an exothermic reaction if not mixed properly.
Molarity Formula
Molarity is a key concept in solution chemistry that describes the concentration of a solution. Defined as the number of moles of solute per liter of solution, molarity is expressed as moles per liter (mol/L), often abbreviated as M.
The formula to calculate molarity is:
\[\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]
To find how much volume of a solution is needed for a specific concentration, you can rearrange the formula:
\[\text{Volume (L)} = \frac{\text{Moles of solute}}{\text{Molarity}}\]
In practice, understanding and using the molarity formula allows chemists to prepare solutions of exact concentrations needed for experiments or production processes. It's vital for ensuring reactions occur under controlled and predictable conditions.

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Most popular questions from this chapter

In some laboratory analyses, the preferred technique is to dissolve a sample in an excess of acid or base and then "back-titrate" the excess with a standard base or acid. This technique is used to assess the purity of a sample of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\). Suppose you dissolve a 0.475 -g sample of impure \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in aqueous \(\mathrm{KOH}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The NH ated in the reaction is distilled from the solution into a flask containing \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCl. The ammonia reacts with the acid to produce \(\mathrm{NH}_{4} \mathrm{Cl},\) but not all of the \(\mathrm{HCl}\) is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes \(11.1 \mathrm{mL}\) of \(0.121 \mathrm{M} \mathrm{NaOH}\). What is the weight percent of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in the \(0.475-\mathrm{g}\) sample?

What mass of lime, CaO, can be obtained by heating \(125 \mathrm{kg}\) of limestone that is \(95.0 \%\) by mass \(\mathrm{CaCO}_{3} ?\) $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$

Aqueous solutions of iron(II) chloride and sodium sulfide react to form iron(11)sulfide and sodium chloride. (a) Write the balanced equation for the reaction. (b) If you combine \(40 .\) g each of \(\mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{FeCl}_{2}\), what is the limiting reactant? (c) What mass of FeS is produced? (d) What mass of NasS or FeCl, remains after the reaction? (e) What mass of \(\mathrm{FeCl}_{2}\) is required to react completely with \(40 . \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{S} ?\)

ATOM ECONOMY: One type of reaction used in the chemical industry is a substitution, where one atom or group is exchanged for another. In this reaction, an alcohol, 1-butanol, is transformed into 1 -bromobutane by substituting Br for the -OH group in the presence of sulfuric acid. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{NaBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow\) $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O} $$ Calculate the \(\%\) atom economy for the desired product, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}.\)

Gold can be dissolved from gold-bearing rock by treating the rock with sodium cyanide in the presence of oxygen. \(4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq}) $$ (a) Name the oxidizing and reducing agents in this reaction. What has been oxidized, and what has been reduced? (b) If you have exactly one metric ton ( 1 metric ton \(=1000 \mathrm{kg})\) of gold-bearing rock, what volume of \(0.075 \mathrm{M} \mathrm{NaCN},\) in liters, do you need to extract the gold if the rock is \(0.019 \%\) gold?

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