If a 0.890 m aqueous solution freezes at -3.00 °C, what is the van't Hoff factor, i, of the solute? Consult the table of Kf values. i = Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Kf value* Normal freezing K value Normal boiling (°C/ m) Solvent Formula (°C/ m) point ( °C) point ( °C) H20 C6H6 water 1.86 0.00 0.512 100.00 benzene 5.12 5.49 2.53 80.1 cyclohexane C 6 H 12 C2H60 20.8 6.59 2.92 80.7 ethanol 1.99 |-117.3 1.22 78.4 carbon tetrachloride CCI 4 camphor 29.8 -22.9 5.03 76.8 С 10 Н 16 0 37.8 176

If a 0.890 m aqueous solution freezes at -3.00 °C, what is the van't Hoff factor, i, of the solute? Consult the table of Kf values. i = Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Kf value* Normal freezing K value Normal boiling (°C/ m) Solvent Formula (°C/ m) point ( °C) point ( °C) H20 C6H6 water 1.86 0.00 0.512 100.00 benzene 5.12 5.49 2.53 80.1 cyclohexane C 6 H 12 C2H60 20.8 6.59 2.92 80.7 ethanol 1.99 |-117.3 1.22 78.4 carbon tetrachloride CCI 4 camphor 29.8 -22.9 5.03 76.8 С 10 Н 16 0 37.8 176

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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The increase in the boiling point of a solution is given by the equation: ДТ — Кті Where: AT = boiling point elevation K = boiling point elevation constant m = molality of the solution i van't Hoff factor NOTE: Write AT as dT (with no spaces) for the answers below. In order to solve for the van't Hoff factor, i, you must multiply both sides of the equation by the same expression: ДТ х AT x= Kmi x The resulting equation is:
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