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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?

Short Answer

Expert verified
The concentration of the diluted solution is 0.082 M.

Step by step solution

01

Identify the Knowns

First identify all the known values. The volume of the initial solution (V1) is given as 10.00 mL or 0.01 L since the volumes in this formula must be in Liters. The molarity of the initial solution (M1) is given as 2.05 M. The final volume after dilution (V2) is given as 250.0 mL or 0.25 L. The goal is to find the molarity of the solution after dilution (M2), which is currently unknown.
02

Apply Dilution Formula

To determine the molarity of the diluted solution, use the molarity dilution equation: M1V1 = M2V2. Plug all known values into the equation: (2.05 M)*(0.01 L) = M2* (0.25 L).
03

Solve for the Unknown

Next, rearrange the equation by dividing both sides by 0.25 L to solve for M2, the molarity of the diluted solution. This leads to the following calculation: M2 = (2.05 M * 0.01 L) / 0.25 L.
04

Calculate Molarity of the Diluted Solution

Perform the calculation in Step 3, leading to a result: M2 = 0.082 M. This is the molarity of the diluted potassium nitrate solution.

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Key Concepts

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

Molarity
The concept of molarity is a fundamental aspect of chemistry. Molarity (M) refers to the concentration of a solution and is defined as the number of moles of solute per liter of solution. It's a way to express how much of a given substance is present in a certain volume of liquid. For example, if you have a solution labeled as `2.05 M KNO3`, it means that there are 2.05 moles of potassium nitrate dissolved in every liter of the solution. Understanding molarity is crucial in many areas of chemistry, including reaction stoichiometry, where knowing the exact concentration allows for the precise calculation of reactants and products. Molarity is typically used in the context of liquids where substances are dissolved in water or another solvent, making it important for laboratory work and industrial applications.
Volume Conversion
Volume conversion is an essential step in many chemical calculations, particularly when you are working with different units. In chemistry, volume is often expressed in milliliters (mL) or liters (L). However, most equations, like the dilution equation used in our example, require the volume to be in liters.
  • 1 liter (L) is equal to 1000 milliliters (mL).
  • To convert from milliliters to liters, divide the number of milliliters by 1000.
For instance, when given a volume of `250.0 mL`, converting to liters involves dividing by 1000, yielding `0.25 L`. This conversion is crucial because using the correct unit ensures that the values work properly in calculations, reducing the risk of errors.
Dilution Formula
The dilution formula is a simple yet powerful tool in chemistry. The formula, expressed as \( M_1V_1 = M_2V_2 \), is used to relate the concentrations and volumes of two solutions before and after dilution. This equation can be particularly useful when trying to determine the final concentration of a solution after adding solvent.Here's a breakdown of what each term represents:
  • \( M_1 \) is the initial molarity (concentration) of the solution.
  • \( V_1 \) is the initial volume of the solution.
  • \( M_2 \) is the final molarity after dilution.
  • \( V_2 \) is the final volume of the solution.
In our original exercise, knowing the initial concentration and volume allows us to apply this formula to determine the new concentration after dilution. When we rearrange to solve for \( M_2 \), the resulting formula becomes \( M_2 = \frac{M_1V_1}{V_2} \). This approach ensures accurate predictions regarding how dilution affects solution concentration, making it an essential concept for mixing solutions in both laboratory and industrial settings.

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Most popular questions from this chapter

The reaction of calcium hydride and water produces calcium hydroxide and hydrogen as products. How many moles of \(\mathrm{H}_{2}(\mathrm{g})\) will be formed in the reaction between \(0.82 \mathrm{mol} \mathrm{CaH}_{2}(\mathrm{s})\) and \(1.54 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ?\)

A small piece of zinc is dissolved in \(50.00 \mathrm{mL}\) of \(1.035 \mathrm{M}\) HCl. At the conclusion of the reaction, the concentration of the \(50.00 \mathrm{mL}\) sample is redetermined and found to be \(0.812 \mathrm{M} \mathrm{HCl} .\) What must have been the mass of the piece of zinc that dissolved? $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

Nitrogen gas, \(\mathrm{N}_{2}\), can be prepared by passing gaseous ammonia over solid copper(II) oxide, \(\mathrm{CuO}\), at high temperatures. The other products of the reaction are solid copper, \(\mathrm{Cu},\) and water vapor. In a certain experiment, a reaction mixture containing \(18.1 \mathrm{g} \mathrm{NH}_{3}\) and \(90.4 \mathrm{g}\) CuO yields \(6.63 \mathrm{g} \mathrm{N}_{2}\). Calculate the percent yield for this experiment.

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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?

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