To prepare a solution that is \(0.50 \mathrm{M} \mathrm{KCl}\) starting with \(100.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{KCl},\) you should \((\mathrm{a})\) add \(20.0 \mathrm{mL}\) of water; (b) add 0.075 g KCl; (c) add 0.10 mol KCl; (d) evaporate \(20.0 \mathrm{mL}\) of water.

Short Answer

Expert verified
The correct option to prepare a \(0.50 M\ \mathrm{KCl}\) solution starting with \(100.0 \mathrm{mL}\) of a \(0.40 M\ \mathrm{KCl}\) solution is to evaporate \(20.0 \mathrm{mL}\) of water.

Step by step solution

01

- Evaluate the first option

In the first option (a), we add \(20.0 \mathrm{mL}\) of water to the initial solution. This increases the volume of the solution, therefore the concentration decreases, which does not achieve the intended result.
02

- Evaluate the second option

In the second option (b), we add 0.075 g KCl to the initial solution. Potassium chloride (KCl) has a molar mass of \(74.55 \mathrm{g/mol}\). So, 0.075 g is approximately 0.001 mol. This increases the amount of solute, but not enough to reach a \(0.50 M\ \mathrm{KCl}\) concentration.
03

- Evaluate the third option

In the third option (c), we add 0.10 mol KCl to the initial solution. This significantly increases the amount of solute, leading to a concentration change. However, adding such a large amount of solute would far exceed the desired concentration.
04

- Evaluate the fourth option

In the fourth option (d), we evaporate \(20.0 \mathrm{mL}\) of water from the initial solution. This reduces the volume of the solution, leading to an increase in concentration. The new volume would be \(100.0 \mathrm{mL} - 20.0 \mathrm{mL} = 80.0 \mathrm{mL}\).\n\nBy substituting the numbers into the formula \(C = n/V\) where \(C\) is the molarity, \(n\) is the number of moles and \(V\) is the volume in liters, we get \n\n\(C = 0.40 \mathrm{mol/L} \times 0.100 \mathrm{L} / 0.080 \mathrm{L} = 0.50 \mathrm{M KCl}\).\n\nSo option (d) is the correct procedure to prepare a \(0.50 M\ \mathrm{KCl}\) solution starting with \(100.0 \mathrm{mL}\) of a \(0.40 M\ \mathrm{KCl}\) solution.

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Key Concepts

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

Molarity Calculations
Understanding molarity is crucial in chemistry, as it provides information on the concentration level in a solution. Essentially, molarity (\( M \)) is defined as the number of moles of solute per liter of solution. It's calculated using the formula: \[ M = \frac{n}{V} \] where \( n \) represents the amount of solute in moles, and \( V \) is the volume of the solution in liters.

To illustrate this, let's consider our textbook example where we want to achieve a solution of \(0.50 M \mathrm{KCl}\). Initially, we have \(100.0 mL\) of a \(0.40 M \mathrm{KCl}\) solution. To reach a molarity of \(0.50 M\), we need to manipulate the number of moles of solute or the volume of the solution. If we decide to add a certain mass of KCl, we first need to convert that mass to moles using the molar mass of KCl. This type of calculation can be tricky because you have to take the volume change into account after adding the solute.

However, by evaporating water, we decrease the volume and increase the concentration without changing the amount of solute—a clear illustration of how molarity can be affected.
Concentration of Solutions
Concentration of a solution is a measure of the amount of solute dissolved in a given quantity of solvent or solution. It's a broad term that includes molarity but also other measurements such as mass percent, volume percent, and molality. Concentration can determine the reactivity, boiling point, and many other properties of the solution.

It's important to remember that when adding more solute, you increase the concentration, assuming the volume remains constant. Alternatively, reducing the volume, as in our exercise through evaporation, also results in a higher molarity. Hence, the method of reaching the desired concentration depends on whether you are working with the solute amount, volume of solution, or both.
Dilution and Evaporation
Dilution and evaporation are two processes that can alter the concentration of a solution. Dilution involves adding more solvent, which increases the volume and lowers the concentration of the solution. In contrast, evaporation removes solvent, usually water, thereby reducing the volume and increasing the concentration of the solution.

These processes are applied in various scenarios in a laboratory or industrial setting. For instance, in our given exercise, to prepare a \(0.50 M \mathrm{KCl}\) solution from a \(0.40 M \mathrm{KCl}\) solution, we opt for evaporating water. Evaporating \(20.0 mL\) from our initial \(100.0 mL\) of solution increases the solution's concentration by decreasing its volume—an effective and practical approach to controlling concentration without the need for additional solute.

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Most popular questions from this chapter

A commercial method of manufacturing hydrogen involves the reaction of iron and steam. $$ 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) $$ (a) How many grams of \(\mathrm{H}_{2}\) can be produced from \(42.7 \mathrm{g}\) Fe and an excess of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (steam)? (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are consumed in the conversion of \(63.5 \mathrm{g}\) Fe to \(\mathrm{Fe}_{3} \mathrm{O}_{4} ?\) (c) If \(14.8 \mathrm{g} \mathrm{H}_{2}\) is produced, how many grams of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) must also be produced?

Which of the following is a \(0.500 \mathrm{M} \mathrm{KCl}\) solution? (a) \(0.500 \mathrm{g} \mathrm{KCl} / \mathrm{mL}\) solution; (b) \(36.0 \mathrm{g} \mathrm{KCl} / \mathrm{L}\) solu- tion; (c) 7.46 mg KCl/mL solution; (d) 373 g KCl in 10.00 L solution

A \(1.000 \mathrm{g}\) sample of a mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) is analyzed by burning it completely in \(\mathrm{O}_{2}\), yielding \(2.776 \mathrm{g} \mathrm{CO}_{2} .\) What is the percentage by mass of \(\mathrm{CH}_{4}\) in the mixture? (a) \(93 \% ;\) (b) \(82 \% ;\) (c) \(67 \% ;\) (d) \(36 \%\) (e) less than \(36 \%\)

A 10.00 mL sample of \(2.05 \mathrm{M} \mathrm{KNO}_{3}\) is diluted to a volume of \(250.0 \mathrm{mL}\). What is the concentration of the diluted solution?

Lead nitrate and potassium iodide react in aqueous solution to form a yellow precipitate of lead iodide. In one series of experiments, the masses of the two reactants were varied, but the total mass of the two was held constant at \(5.000 \mathrm{g}\). The lead iodide formed was filtered from solution, washed, dried, and weighed. The table gives data for a series of reactions. $$\begin{array}{lll} \hline & \text { Mass of Lead } & \text { Mass of Lead } \\ \text { Experiment } & \text { Nitrate, } g & \text { lodide, } g \\ \hline 1 & 0.500 & 0.692 \\ 2 & 1.000 & 1.388 \\ 3 & 1.500 & 2.093 \\ 4 & 3.000 & 2.778 \\ 5 & 4.000 & 1.391 \\ \hline \end{array}$$ (a) Plot the data in a graph of mass of lead iodide versus mass of lead nitrate, and draw the appropriate curve(s) connecting the data points. What is the maximum mass of precipitate that can be obtained? (b) Explain why the maximum mass of precipitate is obtained when the reactants are in their stoichiometric proportions. What are these stoichiometric proportions expressed as a mass ratio, and as a mole ratio? (c) Show how the stoichiometric proportions determined in part (b) are related to the balanced equation for the reaction.

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