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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: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter13: The Chemistry Of Solutes And Solutions
Section13.7: Colligative Properties Of Solutions
Problem 13.17CE
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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
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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