QUESTION 6 Rubbing alcohol is 70.% (m/v) isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 1 pint (473 mL) container? O A) 70. mL O B) 0.15 mL O C) 680 mL OD) 330 mL QUESTION 7 Click Save and Submit to save and submit. Click Save All Answers to save all answers.
QUESTION 6 Rubbing alcohol is 70.% (m/v) isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 1 pint (473 mL) container? O A) 70. mL O B) 0.15 mL O C) 680 mL OD) 330 mL QUESTION 7 Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Educational Website Content
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**QUESTION 6**
Rubbing alcohol is 70.0% (m/v) isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 1 pint (473 mL) container?
- A) 70. mL
- B) 0.15 mL
- C) 680 mL
- D) 330 mL
**QUESTION 7**
---
**Instructions:**
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
---
**Explanation:**
In this problem, you are asked to calculate the volume of isopropyl alcohol in a given quantity of rubbing alcohol, which is 70.0% isopropyl alcohol by volume.
**Steps to solve:**
1. Identify the given data and required data.
2. Use concentration formula:
\[
\text{Concentration} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100
\]
3. Rearrange the formula to find the volume of the solute (isopropyl alcohol):
\[
\text{Volume of solute} = \text{Concentration} \times \text{Volume of solution}
\]
4. Plug in the values into the equation:
\[
\text{Volume of solute} = 70.0\% \times 473 \text{ mL}
\]
\[
\text{Volume of solute} = 0.70 \times 473 \text{ mL}
\]
\[
\text{Volume of solute} = 331.1 \text{ mL}
\]
Therefore, the correct answer is closest to option D) 330 mL.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c1eccff-4d99-45a2-a9f3-7cdc9dec6dc3%2F1182c304-bad5-434d-a305-b99b6a8986f8%2Fmpxqwm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Website Content
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**QUESTION 6**
Rubbing alcohol is 70.0% (m/v) isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 1 pint (473 mL) container?
- A) 70. mL
- B) 0.15 mL
- C) 680 mL
- D) 330 mL
**QUESTION 7**
---
**Instructions:**
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
---
**Explanation:**
In this problem, you are asked to calculate the volume of isopropyl alcohol in a given quantity of rubbing alcohol, which is 70.0% isopropyl alcohol by volume.
**Steps to solve:**
1. Identify the given data and required data.
2. Use concentration formula:
\[
\text{Concentration} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100
\]
3. Rearrange the formula to find the volume of the solute (isopropyl alcohol):
\[
\text{Volume of solute} = \text{Concentration} \times \text{Volume of solution}
\]
4. Plug in the values into the equation:
\[
\text{Volume of solute} = 70.0\% \times 473 \text{ mL}
\]
\[
\text{Volume of solute} = 0.70 \times 473 \text{ mL}
\]
\[
\text{Volume of solute} = 331.1 \text{ mL}
\]
Therefore, the correct answer is closest to option D) 330 mL.
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