/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A Suppose you have \(100.00 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A Suppose you have \(100.00 \mathrm{mL}\) of a solution of a dye and transfer \(2.00 \mathrm{mL}\) of the solution to a \(100.00-\mathrm{mL}\) volumetric flask. After adding water to the \(100.00 \mathrm{mL}\) mark, you take 5.00 mL. of that solution and again dilute to \(100.00 \mathrm{mL}\). If you find the dye concentration in the final diluted sample is \(0.000158 \mathrm{M},\) what was the dye concentration in the original solution?

Short Answer

Expert verified
The original concentration was \(0.158 \text{ M}\).

Step by step solution

01

Determine concentration after first dilution

In the first dilution, we transferred 2.00 mL of the original solution to a 100.00 mL volumetric flask. Use the dilution formula \( C_1V_1 = C_2V_2 \). Here, \( V_1 = 2.00 \) mL and \( V_2 = 100.00 \) mL. Let \( C_1 \) be the initial concentration and \( C_2 \) be the concentration after the first dilution: \[ C_2 = \frac{C_1 \times 2.00}{100.00} \].
02

Determine concentration after second dilution

Take 5.00 mL from the first diluted solution and add it to another 100.00 mL flask: \( V_1 = 5.00 \) mL and \( V_2 = 100.00 \) mL. The final concentration \( C_f \) after this dilution is given as \(0.000158\) M. Therefore, \[ C_2 \times 5.00 = 0.000158 \times 100.00 \]. Solve for \( C_2 \): \[ C_2 = \frac{0.000158 \times 100.00}{5.00} = 0.00316 \text{ M} \].
03

Solve for the original concentration

Now that we have \( C_2 = 0.00316 \text{ M} \) from the first dilution, use it to find \( C_1 \): \[ C_1 = \frac{C_2 \times 100.00}{2.00} = \frac{0.00316 \times 100.00}{2.00} = 0.158 \text{ M} \].
04

Final Answer

The dye concentration in the original solution was \(0.158 \text{ M}\).

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Key Concepts

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

Concentration Calculation
When dealing with solutions, determining the concentration is crucial. Concentration essentially tells us how much solute is present in a given volume of solution. It's usually measured in units of Molarity (M), which is the moles of solute per liter of solution.

In our example, we started with an unknown concentration of dye in a solution. To find this, we performed a series of dilutions. Each dilution made it easier to track how the concentration changed. Solving this involved back-tracking from the known final concentration to establish the original concentration of the dye.

In these calculations, two main principles are key:
  • Moles of solute before and after dilution remain the same.
  • The relationship between volumes and concentrations in dilutions, governed by the dilution formula.

In the exercise, we calculated the concentration ( C_1 ) by rearranging the formula for each dilution stage. We found the dye concentration started as 0.158 M, which matches our understanding of concentration and dilution effects.
Dilution Formula
The dilution formula is a powerful tool for adjusting solution concentrations. It is expressed as \(C_1V_1 = C_2V_2\), where:
  • \(C_1\) represents the initial concentration of the solution.
  • \(V_1\) is the volume of the original concentrated solution.
  • \(C_2\) is the concentration after dilution.
  • \(V_2\) is the final total volume after dilution.

This equation relies on the idea that the amount of solute remains constant before and after the dilution. By simply rearranging this equation, you can solve for any unknown variable as long as you have the other three.

In our example exercise, we moved through two stages of dilution. Each time, using \(C_1V_1 = C_2V_2\) helped us to systematically find intermediate and final concentrations, leading back to the original concentration of the dye. Understanding how to manipulate this equation is a fundamental skill that saves time and errors in lab work.
Volumetric Flask
A volumetric flask is a piece of laboratory glassware known for its precision and accuracy. It is specifically designed to measure one exact volume, which helps ensure solutions are made to accurate concentrations. Volumetric flasks are used in this exercise to create dilutions by filling them up to a specific volume marking.

Here's how they work:
  • Choose a volumetric flask of the desired final volume.
  • Add the solute or concentrated solution to the flask.
  • Carefully add solvent, usually water, until the bottom of the meniscus sits on the marked line.

In our example, volumetric flasks of 100.00 mL played a crucial role in accurately diluting the original dye solution in two steps. By carefully transferring specific volumes to these flasks and topping them up to the calibration mark, we achieved precise dilutions needed for correct concentration calculations.

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Most popular questions from this chapter

A You mix 25.0 mL of 0.234 M FeC 1 , with 42.5 mL of \(0.453 \mathrm{M} \mathrm{NaOH}\) (a) What mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) (in grams) will precipitate from this reaction mixture? (b) One of the reactants \(\left(\mathrm{FeCl}_{3} \text { or } \mathrm{NaOH}\) ) is present \right. in a stoichiometric excess. What is the molar concentration of the excess reactant remaining in solution after \(\mathrm{Fe}(\mathrm{OH})\), has been precipitated?

What volume of \(0.123 \mathrm{M} \mathrm{NaOH},\) in milliliters, contains \(25.0 \mathrm{g}\) of \(\mathrm{NaOH} ?\)

A Commercial sodium "hydrosulfite" is \(90.1 \%\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} .\) The sequence of reactions used to prepare the compound is $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s}) $$ \(\mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})\) (a) What mass of pure \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) can be prepared from \(125 \mathrm{kg}\) of \(\mathrm{Zn}, 500 . \mathrm{g}\) of \(\mathrm{SO}_{2},\) and an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) (b) What mass of the commercial product would contain the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) produced using the amounts of reactants in part (a)?

Identify the ions that exist in each aqueous solution, and specify the concentration of each ion. (a) \(0.25 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (b) \(0.123 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(0.056 \mathrm{M} \mathrm{HNO}_{3}\)

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical las it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers \(1 .\) One way to make the compound is called the "chlorohydrin route." $$ \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Another route is the modern catalytic reaction. $$ \mathrm{C}_{2} \mathrm{H}_{4}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} $$ (a) Calculate the \(\%\) atom economy for the production of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) in each of these reactions. Which is the more efficient method? (b) What is the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) if \(867 \mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

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