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Chapter 9 Exercise 13 Using % m/v as a Conversion Factor

Chapter 9 Exercise 13: Using % m/v as a Conversion Factor

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Introduction

This video covers Chapter 9, Exercise 13, focusing on how to use mass/volume percent (% m/v) as a conversion factor to determine the volume of a solution required to deliver a specific mass of solute. The example involves calculating the volume of a glucose solution needed to provide a patient with 100 grams of glucose.

Main Points
  1. Identifying Key Information

- Problem Statement: A patient needs 100 grams of glucose over 12 hours. Calculate the volume of a 5% mass/volume glucose solution required.

- Key Data: 

     - 100 grams of glucose (solute)
     - 5% mass/volume (m/v) glucose solution
     - Volume of solution needed (liters)
  1. Understanding % m/v

- Definition: 5% m/v means there are 5 grams of glucose in every 100 milliliters of solution.

- Formula:

     \[
     \% \text{ m/v} = \left( \frac{\text{grams of solute}}{\text{milliliters of solution}} \right) \times 100
     \]
  1. Step-by-Step Conversion

- Step 1: Write the given mass of glucose:

     \[
     100 \text{ grams of glucose}
     \]

- Step 2: Use the % m/v conversion factor to find the volume in milliliters:

     \[
     100 \text{ grams} \times \left( \frac{100 \text{ milliliters of solution}}{5 \text{ grams of glucose}} \right) = 2000 \text{ milliliters of solution}
     \]

- Step 3: Convert milliliters to liters:

     \[
     2000 \text{ milliliters} \times \left( \frac{1 \text{ liter}}{1000 \text{ milliliters}} \right) = 2 \text{ liters of solution}
     \]
  1. Calculation

- The patient needs 2 liters of the 5% m/v glucose solution to receive 100 grams of glucose.

Conclusion

By using % m/v as a conversion factor, one can easily determine the volume of solution needed to deliver a specific mass of solute. In this example, 2 liters of a 5% m/v glucose solution are required to provide a patient with 100 grams of glucose.

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