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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

A laboratory method of preparing \(\mathrm{O}_{2}(\mathrm{g})\) involves the decomposition of \(\mathrm{KClO}_{3}(\mathrm{s})\) $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) can be produced by the decomposition of \(32.8 \mathrm{g} \mathrm{KClO}_{3} ?\) (b) How many grams of \(\mathrm{KClO}_{3}\) must decompose to produce \(50.0 \mathrm{g} \mathrm{O}_{2} ?\) (c) How many grams of KCl are formed, together with \(28.3 \mathrm{g} \mathrm{O}_{2},\) in the decomposition of \(\mathrm{KClO}_{3} ?\)

What volume of \(0.149 \mathrm{M} \mathrm{HCl}\) must be added to \(1.00 \times 10^{2} \mathrm{mL}\) of \(0.285 \mathrm{M} \mathrm{HCl}\) so that the resulting solution has a molarity of \(0.205 \mathrm{M} ?\) Assume that the volumes are additive.

The reaction of potassium superoxide, \(\mathrm{KO}_{2}\), is used in life- support systems to replace \(\mathrm{CO}_{2}(\mathrm{g})\) in expired air with \(\mathrm{O}_{2}(\mathrm{g}) .\) The unbalanced chemical equation for the reaction is given below. $$\mathrm{KO}_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})$$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) are produced by the reaction of \(156 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) with excess \(\mathrm{KO}_{2}(\mathrm{s}) ?\) (b) How many grams of \(\mathrm{KO}_{2}(\mathrm{s})\) are consumed per \(100.0 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) removed from expired air? (c) How many \(\mathrm{O}_{2}\) molecules are produced per milligram of \(\mathrm{KO}_{2}\) consumed?

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

It is often difficult to determine the concentration of a species in solution, particularly if it is a biological species that takes part in complex reaction pathways. One way to do this is through a dilution experiment with labeled molecules. Instead of molecules, however, we will use fish. An angler wants to know the number of fish in a particular pond, and so puts an indelible mark on 100 fish and adds them to the pond's existing population. After waiting for the fish to spread throughout the pond, the angler starts fishing, eventually catching 18 fish. Of these, five are marked. What is the total number of fish in the pond?

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