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The normal freezing point of a certain liquid X is 5.10 °C, but when 0.12 kg of alanine (C¿H,NO,) are dissolved in 800. g of X the solution freezes at 0.9 °C instead. Use this information to calculate the molal freezing point depression constant K, of X. Round your answer to 2 significant digits. °C-kg K, - 0 mol

The normal freezing point of a certain liquid X is 5.10 °C, but when 0.12 kg of alanine (C¿H,NO,) are dissolved in 800. g of X the solution freezes at 0.9 °C instead. Use this information to calculate the molal freezing point depression constant K, of X. Round your answer to 2 significant digits. °C-kg K, - 0 mol

Chemistry
Chemistry
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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
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Using the Kt and Kb equations with electrolytes The normal freezing point of a certain liquid X is 5.10 °C, but when 0.12 kg of alanine (C,H,NO,) are dissolved in 800. g of X the solution freezes at 0.9 °C instead. Use this information to calculate the molal freezing point depression constant K, of X. Round your answer to 2 significant digits. °C-kg K, = mol Explanation Check O 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Cent FEB 3. étv A
Transcribed Image Text:Using the Kt and Kb equations with electrolytes The normal freezing point of a certain liquid X is 5.10 °C, but when 0.12 kg of alanine (C,H,NO,) are dissolved in 800. g of X the solution freezes at 0.9 °C instead. Use this information to calculate the molal freezing point depression constant K, of X. Round your answer to 2 significant digits. °C-kg K, = mol Explanation Check O 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Cent FEB 3. étv A
Expert Solution
Step 1

given, the mass of C3H7NO2 = 0.12kg = 0.12 ×103 g

mass of solvent X = 800. g = 0.800 kg

molar mass of C3H7NO2 = 89 g mol-1

freezing point of liquid X = 5.10 oC

freezing point of solution = 0.9 oC

 

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