What is the percent by mass of an aqueous solution that is 2.7m ammonium sulfate?

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Question:**

What is the percent by mass of an aqueous solution that is 2.7m ammonium sulfate?

**Explanation:**

This question requires the calculation of the percent by mass of a solution, given its molality. The solution has a molality of 2.7 mol/kg, which means there are 2.7 moles of ammonium sulfate per kilogram of water. To find the percent by mass, you need to calculate the mass of the ammonium sulfate in the solution and then express it as a percentage of the total mass of the solution. 

Steps:

1. Calculate the molar mass of ammonium sulfate \((\text{NH}_4\) \(2\text{SO}_4)\).
2. Use the molality to find the mass of ammonium sulfate.
3. Calculate the total mass of the solution.
4. Use the formula for percent by mass:

\[
\text{Percent by mass} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 100
\]

This calculation is fundamental in chemistry to understand the concentration of solutions.
Transcribed Image Text:**Question:** What is the percent by mass of an aqueous solution that is 2.7m ammonium sulfate? **Explanation:** This question requires the calculation of the percent by mass of a solution, given its molality. The solution has a molality of 2.7 mol/kg, which means there are 2.7 moles of ammonium sulfate per kilogram of water. To find the percent by mass, you need to calculate the mass of the ammonium sulfate in the solution and then express it as a percentage of the total mass of the solution. Steps: 1. Calculate the molar mass of ammonium sulfate \((\text{NH}_4\) \(2\text{SO}_4)\). 2. Use the molality to find the mass of ammonium sulfate. 3. Calculate the total mass of the solution. 4. Use the formula for percent by mass: \[ \text{Percent by mass} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 100 \] This calculation is fundamental in chemistry to understand the concentration of solutions.
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