Skip to main content
Pearson+ LogoPearson+ Logo
Ch.9 Solutions
McMurry - Fundamentals of GOB 8th Edition
McMurry8th EditionFundamentals of GOBISBN: 9780134015187Not the one you use?Change textbook
All textbooksMcMurry 8th EditionCh.9 SolutionsProblem 48
Chapter 9, Problem 48

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

Verified step by step guidance
1
Step 1: Recall the formula for mass/volume percent concentration: \( \text{Mass/Volume Percent} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100 \).
Step 2: For part (a), calculate the mass of KCl. Use the molar mass of KCl (\( 39.10 \ \text{g/mol} \) for K and \( 35.45 \ \text{g/mol} \) for Cl) to find the mass of 0.078 mol of KCl: \( \text{Mass of KCl} = \text{Moles of KCl} \times \text{Molar Mass of KCl} \).
Step 3: Substitute the calculated mass of KCl and the given volume of solution (75 mL) into the mass/volume percent formula: \( \text{Mass/Volume Percent} = \frac{\text{Mass of KCl (g)}}{75 \ \text{mL}} \times 100 \).
Step 4: For part (b), calculate the mass of sucrose (C₁₂H₂₂O₁₁). Use the molar mass of sucrose (\( 12 \times 12.01 \ \text{g/mol} + 22 \times 1.008 \ \text{g/mol} + 11 \times 16.00 \ \text{g/mol} \)) to find the mass of 0.044 mol of sucrose: \( \text{Mass of sucrose} = \text{Moles of sucrose} \times \text{Molar Mass of sucrose} \).
Step 5: Substitute the calculated mass of sucrose and the given volume of solution (380 mL) into the mass/volume percent formula: \( \text{Mass/Volume Percent} = \frac{\text{Mass of sucrose (g)}}{380 \ \text{mL}} \times 100 \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

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

Mass/Volume Percent Concentration

Mass/volume percent concentration is a way to express the concentration of a solution, defined as the mass of solute (in grams) per 100 mL of solution. It is calculated using the formula: (mass of solute in grams / volume of solution in mL) × 100. This measurement is particularly useful in chemistry for understanding how concentrated a solution is.
Recommended video:
Guided course
00:52
Percent Concentrations Concept 1

Molarity and Moles

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. To find the mass of a solute from its moles, one can use the molar mass, which is the mass of one mole of that substance. Understanding how to convert between moles and grams is essential for calculating mass/volume percent concentration.
Recommended video:
Guided course
00:53
Molarity

Dilution and Solution Preparation

Dilution refers to the process of reducing the concentration of a solute in a solution, typically by adding more solvent. When preparing solutions, it is important to know the final volume and the amount of solute needed to achieve the desired concentration. This concept is crucial for accurately calculating mass/volume percent concentrations in various scenarios.
Recommended video:
Guided course
01:03
Dilutions
Related Practice
Textbook Question

The solubility of CO2 gas in water is 0.15 g/100 mL at a CO2 pressure of 760 mmHg.

b. An atmospheric concentration of 380 ppm, CO2 corresponds to a partial pressure of 0.00038 atm. What percentage of the CO2 originally dissolved in the solution in part (a) remains in solution after the soft drink reaches equilibrium with the ambient atmosphere?

1118
views
Textbook Question

How is volume/volume percent concentration defined and for what types of solutions is it typically used?

2317
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?

1637
views
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)

964
views
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?

2038
views
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?

1107
views