Consider the solubilities of a particular solute at two different temperatures. Temperature (C) Solubility (g/100 g H,O) 20.0 36.2 30.0 67.7 Suppose a saturated solution of this solute was made using 44.0 g H,0 at 20.0 °C. How much more solute can be added if the temperature is increased to 30.0 °C? mass: g

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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:**  
\[
\text{mass} = 13.86 \, \text{g}
\]
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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