What is the boiling point of a solution of 63.5 g of iodine, 12, in 500.0 g of chloroform, CHCl3? Assume the iodine is nonvolatile and that the solution is ideal. The kb for chloroform is 3.88°C/m and the normal boiling point is 61.3°C. BP = [?] °C Boiling Point (°C) Enter
What is the boiling point of a solution of 63.5 g of iodine, 12, in 500.0 g of chloroform, CHCl3? Assume the iodine is nonvolatile and that the solution is ideal. The kb for chloroform is 3.88°C/m and the normal boiling point is 61.3°C. BP = [?] °C Boiling Point (°C) Enter
Chemistry
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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
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![**Boiling Point Elevation Calculation Example**
**Problem Statement:**
What is the boiling point of a solution of 63.5 g of iodine, I₂, in 500.0 g of chloroform, CHCl₃? Assume the iodine is nonvolatile and that the solution is ideal.
The \( k_b \) for chloroform is 3.88°C/m and the normal boiling point is 61.3°C.
**Solution:**
To find the boiling point of the solution, use the formula for boiling point elevation:
\[ \Delta T_b = k_b \times m \]
where:
- \( \Delta T_b \) = boiling point elevation
- \( k_b \) = ebullioscopic constant (for chloroform)
- \( m \) = molality of the solution
First, calculate the molality ( \(m\) ) of the solution:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
Calculate the moles of iodine:
- Molar mass of iodine (I₂) = 2 x 126.90 g/mol = 253.8 g/mol
- Moles of iodine = \(\frac {63.5 \text{ g}}{253.8 \text{ g/mol}} ≈ 0.250 \text{ mol}\)
Calculate the mass of solvent in kilograms:
- Mass of chloroform in kg = \(\frac{500.0 \text{ g}}{1000 \text{ g/kg}} = 0.500 \text{ kg}\)
Now, calculate the molality:
\[ m = \frac{0.250 \text{ mol}}{0.500 \text{ kg}} = 0.500 \text{ m} \]
Next, use the boiling point elevation equation:
\[ \Delta T_b = 3.88 \text{ °C/m} \times 0.500 \text{ m} = 1.94 \text{ °C} \]
Finally, add the boiling point elevation to the normal boiling point:
\[ \text{Boiling Point (BP)} = 61.3°C + 1.94°C = 63.24°C \]
**Boiling Point (BP) = 63.24°C**
In the given interactive text box, users can enter the calculated boiling point (63](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07a40a8c-2023-4125-afaf-f3a8d976c8b9%2F58a59df9-b838-4cce-b9a2-2e55a2155ca2%2Fc6ue57g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Boiling Point Elevation Calculation Example**
**Problem Statement:**
What is the boiling point of a solution of 63.5 g of iodine, I₂, in 500.0 g of chloroform, CHCl₃? Assume the iodine is nonvolatile and that the solution is ideal.
The \( k_b \) for chloroform is 3.88°C/m and the normal boiling point is 61.3°C.
**Solution:**
To find the boiling point of the solution, use the formula for boiling point elevation:
\[ \Delta T_b = k_b \times m \]
where:
- \( \Delta T_b \) = boiling point elevation
- \( k_b \) = ebullioscopic constant (for chloroform)
- \( m \) = molality of the solution
First, calculate the molality ( \(m\) ) of the solution:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
Calculate the moles of iodine:
- Molar mass of iodine (I₂) = 2 x 126.90 g/mol = 253.8 g/mol
- Moles of iodine = \(\frac {63.5 \text{ g}}{253.8 \text{ g/mol}} ≈ 0.250 \text{ mol}\)
Calculate the mass of solvent in kilograms:
- Mass of chloroform in kg = \(\frac{500.0 \text{ g}}{1000 \text{ g/kg}} = 0.500 \text{ kg}\)
Now, calculate the molality:
\[ m = \frac{0.250 \text{ mol}}{0.500 \text{ kg}} = 0.500 \text{ m} \]
Next, use the boiling point elevation equation:
\[ \Delta T_b = 3.88 \text{ °C/m} \times 0.500 \text{ m} = 1.94 \text{ °C} \]
Finally, add the boiling point elevation to the normal boiling point:
\[ \text{Boiling Point (BP)} = 61.3°C + 1.94°C = 63.24°C \]
**Boiling Point (BP) = 63.24°C**
In the given interactive text box, users can enter the calculated boiling point (63
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