Table of contents

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

9. Solutions

Molarity

GOB Chemistry

9. Solutions

Molarity

Previous problemNext problem

5:27 minutes

Problem 52

Textbook Question

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

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:

Play a video:

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.

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.

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.

Watch next

Master Molarity with a bite sized video explanation from Jules

Start learning

Related Videos

Related Practice

Multiple Choice

Hypernatremia is a medical condition where a patient has high levels of sodium in their blood, and is the result of the body containing too little water. A patient has a measured sodium level of 165 mM. If 30.0 mL of their blood were drawn, what mass (in ng) of sodium would be present?

2.64 grams of an unknown compound was dissolved in water to yield 150 mL of solution. The concentration of the solution was 0.075 M. What was the molecular weight of the substance?

A solution with a final volume of 750.0 mL was prepared by dissolving 30.00 mL of benzene (C₆H₆, density = 0.8787 g/mL) in dichloromethane. Calculate the molarity of benzene in the solution.

The concentration of cholesterol (C₂₇H₄₆O) in blood is approximately 5.0 mM. How many grams of cholesterol are in 250 mL of blood?

1967

views

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 Na₂S₂O₃contain this number of moles?

1527

views

Textbook Question

In a laboratory experiment, a 10.0-mL sample of NaCl solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and NaCl solution is 36.15 g. After heating, the evaporating dish and dry NaCl have a combined mass of 25.50 g.

b. What is the molarity (M) of the NaCl solution?

1773

views

Textbook Question

A solution is prepared with 70.0 g of HNO₃ and 130.0 g of H₂O. The HNO₃ solution has a density of 1.21 g/mL. (9.4)

d. What is the molarity (M) of the solution?

2181

views

Text Block

text-block - 555b0674

<p>Molarity</p>

1243