**Calculating the Mole Fraction of Acetic Acid in a Solution****Problem Statement:**Acetic acid dissolves in water as a weak acid. Calculate the mole fraction of acetic acid in the solution that contains 35.0 g of \( HC_2H_3O_2 \) and 250 mL of water.**Given Data:**- Mass of Acetic Acid (\( HC_2H_3O_2 \)): 35.0 g- Volume of Water (\( H_2O \)): 250 mL- Molar Mass (MM) of \( HC_2H_3O_2 \): 60.052 g/mol- Molar Mass (MM) of \( H_2O \): 18.016 g/mol- Density (d) of \( H_2O \): 0.998 g/mL**Solution:**To find the mole fraction of acetic acid (\( X_{ HC_2H_3O_2 } \)), we need to follow these steps:1. **Calculate the moles of acetic acid:** \[ \text{Moles of } HC_2H_3O_2 = \frac{\text{Mass of } HC_2H_3O_2}{\text{Molar Mass of } HC_2H_3O_2} = \frac{35.0 \text{ g}}{60.052 \text{ g/mol}} \]2. **Calculate the mass of water:** \[ \text{Mass of } H_2O = \text{Volume of } H_2O \times \text{Density of } H_2O = 250 \text{ mL} \times 0.998 \text{ g/mL} \]3. **Calculate the moles of water:** \[ \text{Moles of } H_2O = \frac{\text{Mass of } H_2O}{\text{Molar Mass of } H_2O} \]4. **Calculate the mole fraction of acetic acid:** \[ X_{ HC_2H_3O_2 } = \frac{\text{Moles of } HC_2H_3O_2}{\text{

Mole fraction is the ratio of number of moles of specific compound in solution to the total moles in…

Acetic acid dissolves in water as a weak acid. Calculate the mole fraction of acetic acid in the solution that contains 35.0 g HC2H3O2 and 250 mL of water. MM HC2H3O2 = 60.052 g/mol MM H2O = 18.016 g/mol d H₂O = 0.998 g/mL XHC₂H30₂ = [?]

Acetic acid dissolves in water as a weak acid. Calculate the mole fraction of acetic acid in the solution that contains 35.0 g HC2H3O2 and 250 mL of water. MM HC2H3O2 = 60.052 g/mol MM H2O = 18.016 g/mol d H₂O = 0.998 g/mL XHC₂H30₂ = [?]

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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Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

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**Calculating the Mole Fraction of Acetic Acid in a Solution**

**Problem Statement:**
Acetic acid dissolves in water as a weak acid. Calculate the mole fraction of acetic acid in the solution that contains 35.0 g of \( HC_2H_3O_2 \) and 250 mL of water.

**Given Data:**
- Mass of Acetic Acid (\( HC_2H_3O_2 \)): 35.0 g
- Volume of Water (\( H_2O \)): 250 mL
- Molar Mass (MM) of \( HC_2H_3O_2 \): 60.052 g/mol
- Molar Mass (MM) of \( H_2O \): 18.016 g/mol
- Density (d) of \( H_2O \): 0.998 g/mL

**Solution:**

To find the mole fraction of acetic acid (\( X_{ HC_2H_3O_2 } \)), we need to follow these steps:

1. **Calculate the moles of acetic acid:**
\[
\text{Moles of } HC_2H_3O_2 = \frac{\text{Mass of } HC_2H_3O_2}{\text{Molar Mass of } HC_2H_3O_2} = \frac{35.0 \text{ g}}{60.052 \text{ g/mol}}
\]

2. **Calculate the mass of water:**
\[
\text{Mass of } H_2O = \text{Volume of } H_2O \times \text{Density of } H_2O = 250 \text{ mL} \times 0.998 \text{ g/mL}
\]

3. **Calculate the moles of water:**
\[
\text{Moles of } H_2O = \frac{\text{Mass of } H_2O}{\text{Molar Mass of } H_2O}
\]

4. **Calculate the mole fraction of acetic acid:**
\[
X_{ HC_2H_3O_2 } = \frac{\text{Moles of } HC_2H_3O_2}{\text{

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