Question

Calculate the boiling point and freezing point of water solutions in the following solutes: For dilute solutions, Kb = 0.52°C/M Kf = 1.86°C/M a 1.20 M glycerol, a nonelectrolyte Boiling point = °C Freezing point = °C b 0.500 M CaCl2, a strong electrolyte Boiling point = °C Freezing point = °C

          Calculate the boiling point and freezing point of water solutions in the following solutes:
For dilute solutions,
Kb = 0.52°C/M
Kf = 1.86°C/M
a 1.20 M glycerol, a nonelectrolyte
Boiling point = °C
Freezing point = °C
b 0.500 M CaCl2, a strong electrolyte
Boiling point = °C
Freezing point = °C

Calculate the boiling point and freezing point of water solutions in the following solutes:
For dilute solutions,
Kb = 0.52°C/M
Kf = 1.86°C/M
a 1.20 M glycerol, a nonelectrolyte
Boiling point = °C
Freezing point = °C
b 0.500 M CaCl2, a strong electrolyte
Boiling point = °C
Freezing point = °C

Added by Vanessa T.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sudhir Kumar

04/24/2022

Step 1

20 M glycerol solution: Boiling point: \[ \Delta T_{b} = i \cdot K_{b} \cdot m \] \[ \Delta T_{b} = 1 \cdot 0.52 \cdot 1.20 = 0.624 \] \[ T_{b,solution} = T_{b,solvent} + \Delta T_{b} = 100 + 0.624 = 100.624 \] Freezing point: \[ \Delta T_{f} = i \cdot K_{f} \cdot Show more…

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Calculate the boiling point and freezing point of water solutions in the following solutes: For dilute solutions, Kb = 0.52°C/M Kf = 1.86°C/M 1.20 M glycerol, a nonelectrolyte Boiling point Freezing point 0.500 M CaCl2, a strong electrolyte Boiling point Freezing point