Skip to main content
Ch.9 Solutions
McMurry - Fundamentals of General, Organic, and Biological Chemistry 8th Edition
McMurry8th EditionFundamentals of General, Organic, and Biological ChemistryISBN: 9780134015187Not the one you use?Change textbook
All textbooksMcMurry 8th EditionCh.9 SolutionsProblem 52
Chapter 9, Problem 52

If you had only 23 g of KOH remaining in a bottle, how many milliliters of 10.0% (m/v) solution could you prepare? How many milliliters of 0.25 M solution?

Verified step by step guidance
1
Step 1: Understand the problem. You are tasked with determining the volume of two different solutions that can be prepared using 23 g of KOH. The first solution is a 10.0% (m/v) solution, and the second is a 0.25 M solution. Note that (m/v) stands for mass/volume percent, and molarity (M) is moles of solute per liter of solution.
Step 2: For the 10.0% (m/v) solution, use the formula for percent concentration: \( \text{Percent (m/v)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 \). Rearrange this formula to solve for the volume of solution: \( \text{Volume (mL)} = \frac{\text{mass of solute (g)} \times 100}{\text{Percent (m/v)}} \). Substitute the given values: mass = 23 g and percent = 10.0%.
Step 3: For the 0.25 M solution, first calculate the molar mass of KOH. The molar mass is the sum of the atomic masses of potassium (K), oxygen (O), and hydrogen (H): \( \text{Molar mass of KOH} = 39.10 \text{ g/mol} + 16.00 \text{ g/mol} + 1.01 \text{ g/mol} \). Use this molar mass to convert the given mass of KOH (23 g) into moles: \( \text{Moles of KOH} = \frac{\text{mass of KOH (g)}}{\text{molar mass of KOH (g/mol)}} \).
Step 4: Use the molarity formula to calculate the volume of the 0.25 M solution: \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \). Rearrange this formula to solve for the volume of solution: \( \text{Volume (L)} = \frac{\text{moles of solute}}{\text{Molarity (M)}} \). Substitute the calculated moles of KOH and the given molarity (0.25 M).
Step 5: Convert the volume of the 0.25 M solution from liters to milliliters by using the conversion factor \( 1 \text{ L} = 1000 \text{ mL} \). This will give you the final volume of the 0.25 M solution in milliliters.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mass/Volume Percentage (m/v)

Mass/volume percentage (m/v) is a concentration unit that expresses the mass of solute in grams per 100 milliliters of solution. In this case, a 10.0% (m/v) KOH solution means there are 10 grams of KOH in every 100 mL of solution. This concept is essential for calculating how much solution can be prepared from a given mass of solute.
Recommended video:
Guided course
00:38
Mass Percent Concept

Molarity (M)

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. A 0.25 M KOH solution contains 0.25 moles of KOH in every liter of solution. Understanding molarity is crucial for converting between mass of solute and volume of solution, especially when preparing solutions of different concentrations.
Recommended video:
Guided course
00:53
Molarity

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the conservation of mass. In this context, it helps determine how much of a solution can be made from a specific mass of KOH by using the relationships between mass, volume, and concentration. Mastery of stoichiometric principles is vital for solving problems related to solution preparation.
Recommended video:
Guided course
01:16
Stoichiometry
Related Practice
Textbook Question

What is the mass/volume percent concentration of the following solutions?

a. 0.078 mol KCl in 75 mL of solution

b. 0.044 mol sucrose (C12H22O11) in 380 mL of solution

Textbook Question

Nalorphine, a relative of morphine, is used to combat withdrawal symptoms in heroin users. How many milliliters of a 0.40% (m/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?

Textbook Question

An aqueous solution that contains 285 ppm of potassium nitrate (KNO3) is being used to feed plants in a garden. What volume of this solution is needed to prepare 2.0 L of a solution that is 75 ppm in KNO3?

1
views
Textbook Question

A dilute aqueous solution of boric acid, H3BO3 is often used as an eyewash. How would you prepare 500.0 mL of a 0.50% (m/v) boric acid solution?

Textbook Question

How many moles of each substance are needed to prepare the following solutions?

a. 50.0 mL of 8.0% (m/v) KCl (MW = 74.55 g/mol)

b. 200.0 mL of 7.5% (m/v) acetic acid (MW = 60.05 g/mol)

Textbook Question

Sodium thiosulfate (Na2S2O3) the major component in photographic fixer solution, reacts with silver bromide to dissolve it according to the following reaction:

AgBr(s) + 2 Na2S2O3(aq) → Na3Ag(S2O3)2(aq) + NaBr(aq)

b. How many mL of 0.02 M Na2S2O3 contain this number of moles?