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…

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter11: Solutions

Section: Chapter Questions

Problem 45P

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Question

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: