If a 0.900 m aqueous solution freezes at -2.00 Celsius, what is the van't Hoff factor, i, of the solute? i= **Colligative Properties**Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:| Solvent | Formula | \(K_f\) value* (°C/m) | Normal freezing point (°C) | \(K_b\) value (°C/m) | Normal boiling point (°C) ||----------------------|-----------|-----------------------|----------------------------|----------------------|--------------------------|| Water | H₂O | 1.86 | 0.00 | 0.512 | 100.00 || Benzene | C₆H₆ | 5.12 | 5.49 | 2.53 | 80.1 || Cyclohexane | C₆H₁₂ | 20.8 | 6.59 | 2.92 | 80.7 || Ethanol | C₂H₆O | 1.99 | -117.3 | 1.22 | 78.4 || Carbon tetrachloride | CCl₄ | 29.8 | -22.9 | 5.03 | 76.8 || Camphor | C₁₀H₁₆O | 37.8 | 176 | | |*When using positive \(K_f\) values, assume that \(\Delta T_f\) is the absolute value of the change in temperature. If you would like \(\Delta T_f\) to be negative, you must use negative \(K_f\) values. Either way, the freezing point of the solution should be lower.

Given: Molality of solution = 0.900 m Freezing point of solution = -2.00oC Freezing point of pure…

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If a 0.900 m aqueous solution freezes at -2.00 Celsius, what is the van't Hoff factor, i, of the solute?

Chemistry

10th Edition

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

If a 0.900 m aqueous solution freezes at -2.00 Celsius, what is the van't Hoff factor, i, of the solute?

i=

**Colligative Properties**

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

| Solvent | Formula | \(K_f\) value* (°C/m) | Normal freezing point (°C) | \(K_b\) value (°C/m) | Normal boiling point (°C) |
|----------------------|-----------|-----------------------|----------------------------|----------------------|--------------------------|
| Water | H₂O | 1.86 | 0.00 | 0.512 | 100.00 |
| Benzene | C₆H₆ | 5.12 | 5.49 | 2.53 | 80.1 |
| Cyclohexane | C₆H₁₂ | 20.8 | 6.59 | 2.92 | 80.7 |
| Ethanol | C₂H₆O | 1.99 | -117.3 | 1.22 | 78.4 |
| Carbon tetrachloride | CCl₄ | 29.8 | -22.9 | 5.03 | 76.8 |
| Camphor | C₁₀H₁₆O | 37.8 | 176 | | |

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

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