Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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### Question 19:

A solution of isobutyl rubbing alcohol (C₄H₁₀O) contains 40.0 mL of isobutyl alcohol and 60.0 mL of water. The density of isobutyl alcohol is 0.803 g/mL. What is the molarity of the isobutyl alcohol solution?

**Solution:**

1. **Determine the mass of isobutyl alcohol:**
\[ \text{Mass} = \text{Volume} \times \text{Density} \]
\[ = 40.0 \, \text{mL} \times 0.803 \, \text{g/mL} \]
\[ = 32.12 \, \text{g} \]

2. **Calculate the number of moles of isobutyl alcohol (C₄H₁₀O):**
\[ \text{Molar mass of } \text{C}_{4}\text{H}_{10}\text{O} = (4 \times 12.01) + (10 \times 1.008) + (1 \times 16.00) \, \text{g/mol} \]
\[ = 48.04 + 10.08 + 16.00 \, \text{g/mol} \]
\[ = 74.12 \, \text{g/mol} \]

\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
\[ = \frac{32.12 \, \text{g}}{74.12 \, \text{g/mol}} \]
\[ = 0.433 \, \text{mol} \]

3. **Calculate the total volume of the solution:**
\[ \text{Total Volume} = 40.0 \, \text{mL} + 60.0 \, \text{mL} \]
\[ = 100.0 \, \text{mL} \]
\[ = 0.100 \, \text{L} \]

4. **Calculate the molarity of the isobutyl alcohol solution:**
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
\[ = \frac{0.
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Transcribed Image Text:### Question 19: A solution of isobutyl rubbing alcohol (C₄H₁₀O) contains 40.0 mL of isobutyl alcohol and 60.0 mL of water. The density of isobutyl alcohol is 0.803 g/mL. What is the molarity of the isobutyl alcohol solution? **Solution:** 1. **Determine the mass of isobutyl alcohol:** \[ \text{Mass} = \text{Volume} \times \text{Density} \] \[ = 40.0 \, \text{mL} \times 0.803 \, \text{g/mL} \] \[ = 32.12 \, \text{g} \] 2. **Calculate the number of moles of isobutyl alcohol (C₄H₁₀O):** \[ \text{Molar mass of } \text{C}_{4}\text{H}_{10}\text{O} = (4 \times 12.01) + (10 \times 1.008) + (1 \times 16.00) \, \text{g/mol} \] \[ = 48.04 + 10.08 + 16.00 \, \text{g/mol} \] \[ = 74.12 \, \text{g/mol} \] \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \] \[ = \frac{32.12 \, \text{g}}{74.12 \, \text{g/mol}} \] \[ = 0.433 \, \text{mol} \] 3. **Calculate the total volume of the solution:** \[ \text{Total Volume} = 40.0 \, \text{mL} + 60.0 \, \text{mL} \] \[ = 100.0 \, \text{mL} \] \[ = 0.100 \, \text{L} \] 4. **Calculate the molarity of the isobutyl alcohol solution:** \[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \] \[ = \frac{0.
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