A 100%7:06 PM Wed Jan 29Unanswered •1 attempt left • Due on Jan 300.100 g sample is dissolved in 10.0 g of a solvent (Kf = 20.0 kg °C/mol), If the freezing point of the solution is found to be 1.22°C lower than thefreezing point of pure solvent, the molecular weight of the sample is-g/mol.3280.6101642441.64Submit

Answered: A 100% 7:06 PM Wed Jan 29 Unanswered •1…

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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A 100%
7:06 PM Wed Jan 29
Unanswered •1 attempt left • Due on Jan 30
0.100 g sample is dissolved in 10.0 g of a solvent (Kf = 20.0 kg °C/mol), If the freezing point of the solution is found to be 1.22°C lower than the
freezing point of pure solvent, the molecular weight of the sample is
-g/mol.
328
0.610
164
244
1.64
Submit

Expert Solution

Step 1

Mass of solute = 0.100 g

Mass of solvent = 10.0 g

Depression in freezing point (ΔT_f) = 1.22 °C

freezing point depression constant (K_f) = 20.0 kg. °C/mol

Step 2

Colligative properties are the properties of the solution that depend on the number of moles of solute. These properties are different from the solute particles. These properties include boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure lowering.

Step 3

The change in freezing point is calculated by the formula as shown below: