Question

The experimentally measured freezing point of a 1.30 m aqueous solution of CaCl2 is -5.80°C. The freezing point depression constant for water is Kf = 1.86°C/m. Assume the freezing point of pure water is 0.00°C. a) What is the value of the van't Hoff factor for this solution? b) What is the predicted freezing point if there were no ion clustering in the solution?

          The experimentally measured freezing point of a 1.30 m aqueous solution of CaCl2 is -5.80°C. The freezing point depression constant for water is Kf = 1.86°C/m. Assume the freezing point of pure water is 0.00°C.
a) What is the value of the van't Hoff factor for this solution?
b) What is the predicted freezing point if there were no ion clustering in the solution?
        
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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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The experimentally measured freezing point of a 1.30 m aqueous solution of CaCl2 is -5.80°C. The freezing point depression constant for water is Kf = 1.86°C/m. Assume the freezing point of pure water is 0.00°C. a) What is the value of the van't Hoff factor for this solution? b) What is the predicted freezing point if there were no ion clustering in the solution?
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Transcript

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00:01 To determine the vaugh factor, we'll use the freezing point depression equation, where the change in freezing point is going to be equal to the vaugh factor multiplied by the molality, multiplied by the freezing point constant for the solvent.
00:16 Rearranging this, we get the vaugh factor equal to the change in the freezing point, divided by molality, divided by kf for water.
00:27 The freezing point, or change in freezing point, is negative 5 .80, celsius.
00:33 The molality provided is 1 .30 molal and kf for water is 1 .86.
00:40 Many people express it as a negative so that we don't get a negative eye value...
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