Chapter 16.2, Problem 26LC

Interpretation Introduction

Interpretation: The mass of $\text{K2SO4}$ required to prepare $1500\text{ g}$ of $5.0\text{ \% K2SO4 (m/m)}$ is to be calculated.

Concept introduction: The concentration of a solution is defined as the amount of solute present in the given quantity of solvent. Concentration is usually measured by various terms and which term is used depends upon the type of solution.

Answer to Problem 26LC

The mass of $\text{K2SO4}$ required to prepare $1500\text{ g}$ of $5.0\text{ \% K2SO4 (m/m)}$ is $\text{ 75 g}$ .

Explanation of Solution

The concentration of a solution is defined by various terms such as:

• Molarity.
• Molality.
• Normality.
• Mole fraction.
• Percent by mass.
• Percent by volume.

The concentration of $\text{ K2SO4}$ is given in percent by mass. The percent by mass is defined as the ratio of the mass of the solute to the mass of the solution. The formula to define the percent by mass is as follows:

  $\%(m/m)=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

The data given in the question to find out the mass of $\text{ K2SO4 }$ is as follows:

• The mass of the solution is $1500\text{ g}$ .
• The concentration of the solution is $5\text{ \% (m/m)}$ .

Substitute the values to find out the percent (v/v).

  $\%(m/m)=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

  $\begin{array}{l}5=\frac{\text{Mass of solute}}{1500\text{ g}}\times 100\\ Mass\text{ of solute}=\frac{1500\text{ g}\times 5}{100}\\ \text{Mass of solute = 75 g}\end{array}$

Conclusion

The mass of $\text{K2SO4}$ required to prepare from $1500\text{ g}$ of $5.0\text{ \% K2SO4 (m/m)}$ is $\text{ 75 g}$ .