Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![The table below shows the solubilities of a particular solute at two different temperatures:
| Temperature (°C) | Solubility (g/100 g H₂O) |
|--------------------|----------------------------|
| 20.0 | 36.2 |
| 30.0 | 67.7 |
**Problem Statement:**
Suppose a saturated solution of this solute was made using 44.0 g H₂O at 20.0 °C. How much more solute can be added if the temperature is increased to 30.0 °C?
**Solution:**
1. Calculate the solubility at 20.0 °C for 44.0 g of water:
\[
\text{Solubility} = 36.2 \, \text{g/100g H₂O}
\]
\[
\text{Solute at 20.0 °C in 44.0 g H₂O} = 36.2 \times \frac{44.0}{100} = 15.928 \, \text{g}
\]
2. Calculate the solubility at 30.0 °C for 44.0 g of water:
\[
\text{Solubility} = 67.7 \, \text{g/100g H₂O}
\]
\[
\text{Solute at 30.0 °C in 44.0 g H₂O} = 67.7 \times \frac{44.0}{100} = 29.788 \, \text{g}
\]
3. Determine how much more solute can be added:
\[
\text{Additional solute} = 29.788 - 15.928 = 13.86 \, \text{g}
\]
Thus, 13.86 grams more solute can be added when the temperature is increased to 30.0 °C.
**Mass:**
\[
\text{mass} = 13.86 \, \text{g}
\]](https://content.bartleby.com/qna-images/question/fe9fe32c-b492-41de-b627-5e5dec5bc081/77bd3458-c831-43e5-a432-c68b65d5d012/iotxvcp_processed.jpeg)
Transcribed Image Text:The table below shows the solubilities of a particular solute at two different temperatures:
| Temperature (°C) | Solubility (g/100 g H₂O) |
|--------------------|----------------------------|
| 20.0 | 36.2 |
| 30.0 | 67.7 |
**Problem Statement:**
Suppose a saturated solution of this solute was made using 44.0 g H₂O at 20.0 °C. How much more solute can be added if the temperature is increased to 30.0 °C?
**Solution:**
1. Calculate the solubility at 20.0 °C for 44.0 g of water:
\[
\text{Solubility} = 36.2 \, \text{g/100g H₂O}
\]
\[
\text{Solute at 20.0 °C in 44.0 g H₂O} = 36.2 \times \frac{44.0}{100} = 15.928 \, \text{g}
\]
2. Calculate the solubility at 30.0 °C for 44.0 g of water:
\[
\text{Solubility} = 67.7 \, \text{g/100g H₂O}
\]
\[
\text{Solute at 30.0 °C in 44.0 g H₂O} = 67.7 \times \frac{44.0}{100} = 29.788 \, \text{g}
\]
3. Determine how much more solute can be added:
\[
\text{Additional solute} = 29.788 - 15.928 = 13.86 \, \text{g}
\]
Thus, 13.86 grams more solute can be added when the temperature is increased to 30.0 °C.
**Mass:**
\[
\text{mass} = 13.86 \, \text{g}
\]
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