A 0.250 g sample of the weak acid HA (molar mass = 150.0 g/mol) is dissolved in 50.0 g of water. The freezing point of the resulting solution is -0.072°C. Calculate the value of Ka for this acid. Assume molality equals molarity in this solution. (Kf for water is 1.86°C/m.)

Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter11: Solutions And Colloids
Section: Chapter Questions
Problem 53E: A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at -1.94...
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A 0.250 g sample of the weak acid HA (molar mass = 150.0 g/mol) is dissolved in 50.0 g of water. The freezing point of the resulting solution is -0.072°C. Calculate the value of Ka for this acid. Assume molality equals molarity in this solution. (Kf for water is 1.86°C/m.)
 
 
 
 
 
 
 
Question 34 of 36
A 0.250 g sample of the weak acid HA
(molar mass = 150.0 g/mol) is dissolved in
50.0 g of water. The freezing point of the
resulting solution is -0.072°C. Calculate the
value of Ka for this acid. Assume molality
equals molarity in this solution. (Kf for water
is 1.86°C/m.)
Transcribed Image Text:Question 34 of 36 A 0.250 g sample of the weak acid HA (molar mass = 150.0 g/mol) is dissolved in 50.0 g of water. The freezing point of the resulting solution is -0.072°C. Calculate the value of Ka for this acid. Assume molality equals molarity in this solution. (Kf for water is 1.86°C/m.)
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