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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![**Calculating Osmolarity of Ammonium Nitrate Solution**
In this section, we will calculate the osmolarity of an aqueous solution containing ammonium nitrate.
**Problem Statement:**
*If an aqueous solution is 2.80% (w/v) in ammonium nitrate, NH₄NO₃, what is the osmolarity of the solution?*
**Formula:**
\[ \text{Osmolarity} = \, \boxed{\, \, \text{osmol/L} } \]
### Steps to Calculate Osmolarity:
1. **Understand the Given Data:**
- The concentration of ammonium nitrate in the solution is 2.80% (w/v). This means there are 2.80 grams of NH₄NO₃ in every 100 mL of solution.
2. **Molecular Dissociation:**
- Ammonium nitrate (NH₄NO₃) dissociates into two ions in water, ammonium ion (NH₄⁺) and nitrate ion (NO₃⁻):
\[ \text{NH₄NO₃} \rightarrow \text{NH₄}^+ + \text{NO₃}^- \]
- Each mole of NH₄NO₃ produces a total of 2 osmoles (one of NH₄⁺ and one of NO₃⁻).
3. **Calculate Molality:**
- We first need the molar mass of NH₄NO₃:
\[
\text{Molar Mass of NH₄NO₃} = (14 + 1*4) + (14 + 16*3) = 80.04 \, g/mol
\]
- Convert the weight/volume percentage to molarity (M):
\[
\text{Molarity (M)} = \frac{2.80 \, g}{80.04 \, g/mol \times 0.1\, L} = \frac{2.80 \, g}{8.004 \, g} = 0.35 \, M
\]
4. **Calculate Osmolarity:**
- Since each mole of NH₄NO₃ gives 2 osmoles, we multiply the molarity by the number of particles:
\[
\text{Osm](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a9ce581-7dc0-475d-98b0-a05b944e1333%2F45311a64-85d8-4d2b-8060-06439f42fff3%2F9ik4knp.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating Osmolarity of Ammonium Nitrate Solution**
In this section, we will calculate the osmolarity of an aqueous solution containing ammonium nitrate.
**Problem Statement:**
*If an aqueous solution is 2.80% (w/v) in ammonium nitrate, NH₄NO₃, what is the osmolarity of the solution?*
**Formula:**
\[ \text{Osmolarity} = \, \boxed{\, \, \text{osmol/L} } \]
### Steps to Calculate Osmolarity:
1. **Understand the Given Data:**
- The concentration of ammonium nitrate in the solution is 2.80% (w/v). This means there are 2.80 grams of NH₄NO₃ in every 100 mL of solution.
2. **Molecular Dissociation:**
- Ammonium nitrate (NH₄NO₃) dissociates into two ions in water, ammonium ion (NH₄⁺) and nitrate ion (NO₃⁻):
\[ \text{NH₄NO₃} \rightarrow \text{NH₄}^+ + \text{NO₃}^- \]
- Each mole of NH₄NO₃ produces a total of 2 osmoles (one of NH₄⁺ and one of NO₃⁻).
3. **Calculate Molality:**
- We first need the molar mass of NH₄NO₃:
\[
\text{Molar Mass of NH₄NO₃} = (14 + 1*4) + (14 + 16*3) = 80.04 \, g/mol
\]
- Convert the weight/volume percentage to molarity (M):
\[
\text{Molarity (M)} = \frac{2.80 \, g}{80.04 \, g/mol \times 0.1\, L} = \frac{2.80 \, g}{8.004 \, g} = 0.35 \, M
\]
4. **Calculate Osmolarity:**
- Since each mole of NH₄NO₃ gives 2 osmoles, we multiply the molarity by the number of particles:
\[
\text{Osm
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