If 31.5 g of LiBr are dissolved 350.0 g of water at 20.0 °C in an insulated container, a temperature change is observed. The ∆H of solution of LiBr is -48.8 kJ/mol. Assuming that the specific heat of the solution is 4.184 J/(g°C), and that no heat is gained or lost by the container, what will be the final temperature of the solution?
If 31.5 g of LiBr are dissolved 350.0 g of water at 20.0 °C in an insulated container, a temperature change is observed. The ∆H of solution of LiBr is -48.8 kJ/mol. Assuming that the specific heat of the solution is 4.184 J/(g°C), and that no heat is gained or lost by the container, what will be the final temperature of the solution?
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Question
If 31.5 g of LiBr are dissolved 350.0 g of water at 20.0 °C in an insulated container, a temperature change is observed. The ∆H of solution of LiBr is -48.8 kJ/mol. Assuming that the specific heat of the solution is 4.184 J/(g°C), and that no heat is gained or lost by the container, what will be the final temperature of the solution?
![Below is the transcription of the provided text, suitable for an educational setting:
---
**Calculation Steps:**
1. **Convert Mass to Moles (LiBr):**
\[
\frac{315 \, \text{g LiBr}}{\left(\frac{1 \, \text{mol LiBr}}{87 \, \text{g LiBr}}\right)} = 3.62 \, \text{mol LiBr}
\]
2. **Calculate Energy Change:**
\[
3.62 \, \text{mol LiBr} \left(-48.8 \frac{\text{kJ}}{\text{mol}}\right) = -176.0656 \, \text{kJ}
\]
Convert to Joules:
\[
-176.0656 \, \text{kJ} \left(\frac{1000 \, \text{J}}{1 \, \text{kJ}}\right) = -176065.6 \, \text{J}
\]
3. **Total Mass of Solution:**
\[
380 \, \text{g H}_{2}\text{O} + 315 \, \text{g LiBr} = 695 \, \text{g solution}
\]
4. **Formula for Heat Transfer:**
\[
-176065.6 \, \text{J} = 381.5 \, \text{g} \left(4.184 \frac{\text{J}}{\text{g°C}}\right) (T_f - 20)
\]
5. **Solve for Final Temperature (T_f):**
\[
-176065.6 \, \text{J} = (T_f - 20)
\]
\[
\frac{-176065.6 \, \text{J}}{381.5 \, \text{g} \cdot 4.184 \frac{\text{J}}{\text{g°C}}} = (T_f - 20)
\]
\[
-11047.3 = T_f - 20
\]
\[
-11047.3 + 20 = T_f
\]
\[
T_f = -11027.3 \, \text{°C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05f56260-8a0e-4ddb-8084-229681fee340%2Feefedd08-9676-4e50-8bcb-3476fc6edb8f%2F0s0ypaf_processed.png&w=3840&q=75)
Transcribed Image Text:Below is the transcription of the provided text, suitable for an educational setting:
---
**Calculation Steps:**
1. **Convert Mass to Moles (LiBr):**
\[
\frac{315 \, \text{g LiBr}}{\left(\frac{1 \, \text{mol LiBr}}{87 \, \text{g LiBr}}\right)} = 3.62 \, \text{mol LiBr}
\]
2. **Calculate Energy Change:**
\[
3.62 \, \text{mol LiBr} \left(-48.8 \frac{\text{kJ}}{\text{mol}}\right) = -176.0656 \, \text{kJ}
\]
Convert to Joules:
\[
-176.0656 \, \text{kJ} \left(\frac{1000 \, \text{J}}{1 \, \text{kJ}}\right) = -176065.6 \, \text{J}
\]
3. **Total Mass of Solution:**
\[
380 \, \text{g H}_{2}\text{O} + 315 \, \text{g LiBr} = 695 \, \text{g solution}
\]
4. **Formula for Heat Transfer:**
\[
-176065.6 \, \text{J} = 381.5 \, \text{g} \left(4.184 \frac{\text{J}}{\text{g°C}}\right) (T_f - 20)
\]
5. **Solve for Final Temperature (T_f):**
\[
-176065.6 \, \text{J} = (T_f - 20)
\]
\[
\frac{-176065.6 \, \text{J}}{381.5 \, \text{g} \cdot 4.184 \frac{\text{J}}{\text{g°C}}} = (T_f - 20)
\]
\[
-11047.3 = T_f - 20
\]
\[
-11047.3 + 20 = T_f
\]
\[
T_f = -11027.3 \, \text{°C
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