Calculate the boiling points of a 8.50 m aqueous solution of fructose. Boiling point constants can be found in the list of constant. Tb= In Celsius **Colligative Properties**Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:| Solvent | Formula | \( K_f \) value (\(^\circ \text{C} \cdot \text{kg/mol}\)) | Normal freezing point (\(^\circ \text{C}\)) | \( K_b \) value (\(^\circ \text{C} \cdot \text{kg/mol}\)) | Normal boiling point (\(^\circ \text{C}\)) ||--------------------|----------|---------------------------------|---------------------------------|--------------------------------|---------------------------------|| Water | \( \text{H}_2\text{O} \) | 1.86 | 0.00 | 0.512 | 100.00 || Benzene | \( \text{C}_6\text{H}_6 \) | 5.12 | 5.49 | 2.53 | 80.1 || Cyclohexane | \( \text{C}_6\text{H}_{12} \) | 20.8 | 6.59 | 2.92 | 80.7 || Ethanol | \( \text{C}_2\text{H}_6\text{O} \) | 1.99 | -117.3 | 1.22 | 78.4 || Carbon tetrachloride | \( \text{CCl}_4 \) | 29.8 | -22.9 | 5.03 | 76.8 || Camphor | \( \text{C}_{10}\text{H}_{16}\text{O} \) | 37.8 | 176 | Custom embed | |*When using positive \( K_f \) values, assume that \(\Delta T_f\) is the absolute value of the change in temperature. If you would prefer, \( \Delta T_f \) can be negative and so you must use negative \( K_f \) values. Either way, the freezing point of the solution should be lowered.*

Formula for boiling point: ∆Tb=Kbmwhere,∆Tb= change in boiling pointkb= boiling point constantm=…

Answered: Calculate the boiling points of a 8.50…

Calculate the boiling points of a 8.50 m aqueous solution of fructose. Boiling point constants can be found in the list of constant.

Human Anatomy & Physiology (11th Edition)

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Chapter1: The Human Body: An Orientation

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Question

Calculate the boiling points of a 8.50 m aqueous solution of fructose. Boiling point constants can be found in the list of constant.
Tb= In Celsius

**Colligative Properties**

Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:

| Solvent | Formula | \( K_f \) value (\(^\circ \text{C} \cdot \text{kg/mol}\)) | Normal freezing point (\(^\circ \text{C}\)) | \( K_b \) value (\(^\circ \text{C} \cdot \text{kg/mol}\)) | Normal boiling point (\(^\circ \text{C}\)) |
|--------------------|----------|---------------------------------|---------------------------------|--------------------------------|---------------------------------|
| Water | \( \text{H}_2\text{O} \) | 1.86 | 0.00 | 0.512 | 100.00 |
| Benzene | \( \text{C}_6\text{H}_6 \) | 5.12 | 5.49 | 2.53 | 80.1 |
| Cyclohexane | \( \text{C}_6\text{H}_{12} \) | 20.8 | 6.59 | 2.92 | 80.7 |
| Ethanol | \( \text{C}_2\text{H}_6\text{O} \) | 1.99 | -117.3 | 1.22 | 78.4 |
| Carbon tetrachloride | \( \text{CCl}_4 \) | 29.8 | -22.9 | 5.03 | 76.8 |
| Camphor | \( \text{C}_{10}\text{H}_{16}\text{O} \) | 37.8 | 176 | Custom embed | |

*When using positive \( K_f \) values, assume that \(\Delta T_f\) is the absolute value of the change in temperature. If you would prefer, \( \Delta T_f \) can be negative and so you must use negative \( K_f \) values. Either way, the freezing point of the solution should be lowered.*

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