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What is the final volume, in milliliters, when \(5.00 \mathrm{~mL}\) of each of the following solutions is diluted to provide the given concentration? (9.5) a. \(20.0 \%\) (m/v) NaOH solution to give a \(4.00 \%\) (m/v) \(\mathrm{NaOH}\) solution b. \(0.600 \mathrm{M} \mathrm{NaOH}\) solution to give a \(0.100 \mathrm{M} \mathrm{NaOH}\) solution c. \(16.0 \%\) (m/v) NaOH solution to give a \(2.00 \%\) (m/v) \(\mathrm{NaOH}\) solution

Short Answer

Expert verified
a. 25.0 mLb. 30.0 mLc. 40.0 mL

Step by step solution

01

Understand the Problem

We need to find the final volume in milliliters of a diluted NaOH solution from a given initial concentration and volume.
02

Use the dilution formula

The dilution formula is based on the principle that the amount of solute remains constant before and after dilution. It is expressed as: \[ C_1V_1 = C_2V_2 \]Where: - \(C_1\) is the initial concentration- \(V_1\) is the initial volume- \(C_2\) is the final concentration- \(V_2\) is the final volume
03

Solve part (a)

Given: Initial concentration \(C_1 = 20.0\text{ % (m/v)}\), Initial volume \(V_1 = 5.00\text{ mL}\), Final concentration \(C_2 = 4.00\text{ % (m/v)}\)Using the dilution formula: \[ 20.0 \times 5.00 = 4.00 \times V_2 \]Solve for \(V_2\): \[ V_2 = \frac{20.0 \times 5.00}{4.00} = 25.0\text{ mL} \]
04

Solve part (b)

Given: Initial concentration \(C_1 = 0.600\text{ M}\), Initial volume \(V_1 = 5.00\text{ mL}\), Final concentration \(C_2 = 0.100\text{ M}\)Using the dilution formula: \[ 0.600 \times 5.00 = 0.100 \times V_2 \]Solve for \(V_2\): \[ V_2 = \frac{0.600 \times 5.00}{0.100} = 30.0\text{ mL} \]
05

Solve part (c)

Given: Initial concentration \(C_1 = 16.0\text{ % (m/v)}\), Initial volume \(V_1 = 5.00\text{ mL}\), Final concentration \(C_2 = 2.00\text{ % (m/v)}\)Using the dilution formula: \[ 16.0 \times 5.00 = 2.00 \times V_2 \]Solve for \(V_2\): \[ V_2 = \frac{16.0 \times 5.00}{2.00} = 40.0\text{ mL} \]

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Key Concepts

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

Dilution Formula
The dilution formula is fundamental in chemistry when working with solutions. It simplifies the process of preparing diluted solutions from more concentrated ones. The dilution formula is expressed as: The idea behind the formula is straightforward: the amount of solute (the substance being dissolved) remains constant before and after dilution.
To effectively use the dilution formula, you should be comfortable with identifying and plugging in the initial concentration ( ), initial volume ( ), final concentration ( ), and final volume ( ). Here is the formula written out:
By rearranging this equation, you can solve for the unknown variable, typically the final volume ( ) if you are performing a dilution task: .
To strengthen this understanding, always ensure each variable is correctly identified and substituted back into the formula. This will yield accurate results for each solution scenario.
Concentration
Concentration is a measure of how much solute is present in a given amount of solution. It can be expressed in different units such as percent by mass/volume ( ) or molarity ( ).
To grasp these concepts, let's break it down:
  • Percent (m/v) is expressed as the mass of solute (in grams) per 100 mL of solution. For example, a 20.0% (m/v) NaOH solution means there are 20.0 grams of NaOH in every 100 mL of the solution.
  • Molarity (M) is the number of moles of solute per liter of solution. A 0.600 M NaOH solution means there are 0.600 moles of NaOH in every liter of the solution.

Understanding the units of concentration is crucial when using the dilution formula. Mismatching units can lead to errors, hence ensure the units match throughout the calculation.
When performing dilution calculations, the initial and final concentrations ( and respectively) dictate how much solute stays constant. Consistency in applying these values in the formula improves accuracy and outcome.
Volume
Volume refers to the amount of space that a substance or object occupies. In the context of dilution calculations, it’s the volume of the solution before and after dilution that we are interested in.

    In dilution problems, the initial volume ( ) and final volume ( ) are integral parts of the dilution formula. To find the final volume when diluting a solution:

Use the dilution formula: Solve for the final volume ( ) by rearranging the formula: The volume changes when you dilute a solution to achieve a specific concentration. You apply water or another solvent to the original solution, increasing its volume while keeping the solute amount the same.
This increased volume results in a lower concentration of the solute. Consistently using the correct units for volume (usually milliliters in these problems) is essential to performing accurate calculations.

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Most popular questions from this chapter

In a laboratory experiment, a 15.0 -mL sample of \(\mathrm{KCl}\) solution is poured into an evaporating dish with a mass of \(24.10 \mathrm{~g}\). The combined mass of the evaporating dish and \(\mathrm{KCl}\) solution is \(41.50 \mathrm{~g} .\) After heating, the evaporating dish and dry \(\mathrm{KCl}\) have a combined mass of \(28.28 \mathrm{~g}\). (9.4) a. What is the mass percent (m/m) of the KCl solution? b. What is the molarity (M) of the KCl solution? c. If water is added to \(10.0 \mathrm{~mL}\) of the initial \(\mathrm{KCl}\) solution to give a final volume of \(60.0 \mathrm{~mL},\) what is the molarity of the diluted KCl solution?

Indicate the compartment (A or B) that will increase in volume for each of the following pairs of solutions separated by a semipermeable membrane: $$\begin{array}{ll}{\mathrm{A}} & {\mathrm{B}} \\\\\hline \text { a. } 20 \%(\mathrm{~m} / \mathrm{v}) \text { starch } & 10 \%(\mathrm{~m} / \mathrm{v}) \text { starch } \\\\\text { b. } 10 \%(\mathrm{~m} / \mathrm{v}) \text { albumin } & 2 \%(\mathrm{~m} / \mathrm{v}) \text { albumin } \\\\\text { c. } 0.5 \%(\mathrm{~m} / \mathrm{v}) \text { sucrose } & 5 \%(\mathrm{~m} / \mathrm{v}) \text { sucrose } \\\\\hline\end{array}$$

What is the difference between a \(5.00 \%(\mathrm{~m} / \mathrm{m})\) glucose solution and a \(5.00 \%(\mathrm{~m} / \mathrm{v})\) glucose solution?

Explain the following observations: a. An open can of soda loses its "fizz" faster at room temperature than in the refrigerator. b. Chlorine gas in tap water escapes as the sample warms to room temperature. c. Less sugar dissolves in iced coffee than in hot coffee.

Describe the formation of an aqueous LiBr solution, when solid LiBr dissolves in water.

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