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Given two liters of \(0.496 \mathrm{M} \mathrm{KCl},\) describe how you would use this solution to prepare \(250.0 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{KCl} .\) Give sufficient details so that another student could follow your instructions.

Short Answer

Expert verified
To prepare 250.0 mL of 0.175 M KCl from a 0.496 M KCl solution, this involves diluting 88.31 mL of the original solution with enough water to obtain a total volume of 250 mL.

Step by step solution

01

Setting up the Dilution Equation

Set up the dilution equation using the formula \(M1V1 = M2V2\), where M1 is initial molarity of the solution, V1 is the initial volume of the solution, M2 is final molarity of the solution, and V2 is final volume of the solution.
02

Substitution

Substitute the known values into the equation. The initial molarity (M1) is \(0.496 M\), the final molarity (M2) is \(0.175 M\), and the final volume V2 is \(250.0 mL\). The equation will be now changed to \(0.496 M \cdot V1 = 0.175 M \cdot 250.0 mL\).
03

Solving the Equation

Next, one needs to solve the equation for the unknown V1. Reframing the equation gives \(V1 = (0.175 M \cdot 250.0 mL) / 0.496 M\). As a result, we get that V1 is approximately \(88.31 mL\). This is the volume of \(0.496 M KCl\) solution required to prepare \(250 mL\) of \(0.175 M KCl\) solution.
04

Procedure Description to Perform the Dilution

To perform the dilution, the first step is to measure \(88.31 mL\) of the \(0.496 M KCl\) solution using a suitable measuring device like a graduated cylinder. Transfer the extracted \(0.496 M KCl\) solution to a container with a capacity of at least \(250 mL\). The remaining volume, which is about \(161.69 mL\) (calculated by subtracting \(88.31 mL\) from \(250 mL\)), should be filled up with distilled water. Stir the solution to ensure uniform distribution of the solute. With this procedure, we get \(250 mL\) of \(0.175 M KCl\) solution.

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Key Concepts

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

Molarity
Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute dissolved per liter of solution. Basically, it tells you how many molecules of a substance are present in a certain volume of liquid. Knowing the molarity allows you to understand how concentrated or dilute a solution is. In chemistry, molarity is denoted by the capital letter **M**. For example, if a solution has a molarity of 0.496 M, this means there are 0.496 moles of solute in every liter of the solution.
You can calculate molarity using the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}. \] This concept is essential in performing dilutions, as it helps you change the concentration of a solution while knowing exactly how much solute is present.
When you prepare a solution of a specific molarity, it means you're ready to use it for chemical reactions or experiments with known concentrations. This is why molarity is a crucial concept in both labs and real-world applications.
Dilution Equation
The dilution equation is an essential tool for calculating how to achieve a desired concentration from a more concentrated solution. It follows the formula \( M1V1 = M2V2 \), where \( M1 \) and \( V1 \) are the molarity and volume of the concentrated solution, and \( M2 \) and \( V2 \) are the molarity and volume of the diluted solution.
This equation ensures that the total amount of solute remains constant before and after the dilution. In practical terms, you use the dilution equation to figure out how much of the concentrated solution you need to mix with a certain amount of water or another solvent to reach the desired concentration.
The beauty of this equation is its simplicity and effectiveness. With just a few values, you can quickly determine how to create the new solution. This makes it especially useful in laboratories and industries where specific concentrations are often required.
Graduated Cylinder
A graduated cylinder is a common laboratory tool used to measure the volume of a liquid very accurately. They are marked with lines or graduations that show specific volumes, making it easier to measure out an exact amount of liquid.
They come in various sizes, from small ones that measure only a few milliliters to larger ones that can measure a liter or more. Graduated cylinders are typically made of glass or plastic, ensuring they are resistant to chemicals and relatively easy to clean.
  • When using a graduated cylinder, you should place it on a flat, level surface and view the liquid from eye level. This helps you avoid parallax error, which can cause inaccurate readings.
  • The curved surface of the liquid in the cylinder, called the meniscus, should be read at the lowest point of the curve to ensure accuracy.
Graduated cylinders are essential when preparing solutions, like when you need to measure out a specific volume of liquid for a dilution, ensuring that you achieve the required concentration of your solution.

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Most popular questions from this chapter

Solid silver oxide, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}),\) decomposes at temperatures in excess of \(300^{\circ} \mathrm{C},\) yielding metallic silver and oxygen gas. A 3.13 g sample of impure silver oxide yields \(0.187 \mathrm{g} \mathrm{O}_{2}(\mathrm{g}) .\) What is the mass percent \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample? Assume that \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) is the only source of \(\mathrm{O}_{2}(\mathrm{g}) .\) [Hint: Write a balanced equation for the reaction.]

Write chemical equations to represent the following reactions. (a) Limestone rock (calcium carbonate) is heated (calcined) and decomposes to calcium oxide and carbon dioxide gas. (b) Zinc sulfide ore is heated in air (roasted) and is converted to zinc oxide and sulfur dioxide gas. (Note that oxygen gas in the air is also a reactant.) (c) Propane gas reacts with gaseous water to produce a mixture of carbon monoxide and hydrogen gases. (This mixture, called synthesis gas, is used to produce a variety of organic chemicals.) (d) Sulfur dioxide gas is passed into an aqueous solution containing sodium sulfide and sodium carbonate. The reaction products are carbon dioxide and an aqueous solution of sodium thiosulfate.

Ammonia can be generated by heating together the solids \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2} . \mathrm{CaCl}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are also formed. (a) If a mixture containing \(33.0 \mathrm{g}\) each of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) is heated, how many grams of \(\mathrm{NH}_{3}\) will form? (b) Which reactant remains in excess, and in what mass?

The rocket boosters of the space shuttle Discovery, launched on July \(26,2005,\) used a fuel mixture containing primarily solid ammonium perchlorate, \(\mathrm{NH}_{4} \mathrm{ClO}_{4}(\mathrm{s}),\) and aluminum metal. The unbalanced chemical equation for the reaction is given below. \(\mathrm{Al}(\mathrm{s})+\mathrm{NH}_{4} \mathrm{ClO}_{4}(\mathrm{s}) \longrightarrow\) $$ \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{AlCl}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{N}_{2}(\mathrm{g}) $$ What is the minimum mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) consumed, per kilogram of \(\mathrm{Al}\), by the reaction of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) and Al?[Hint: Balance the elements in the order \(\mathrm{Cl}, \mathrm{H},\) \(\mathrm{O}, \mathrm{Al}, \mathrm{N} .\)]

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)

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