How would the calculated molar mass compare to the actual molar mass if (circle
ID: 1006790 • Letter: H
Question
How would the calculated molar mass compare to the actual molar mass if (circle the best answer): the freezing point of pure t-butyl alcohol was determine to be too low? Calculated molar mass would be larger than actual molar mass Calculated molar mass would be smaller than actual molar mass Calculated molar mass would be the same as the actual molar mass the freezing point of t-butyl alcohol was correct but the freezing point of the mixture was too high? Calculated molar mass would be larger than actual molar mass Calculated molar mass would be smaller than actual molar mass Calculated molar mass would be the same as the actual molar mass water had contaminated the sample of t-butyl alcohol after measuring the pure freezing point but before adding the unknown solute? Calculated molar mass would be larger than actual molar mass Calculated molar mass would be smaller than actual molar mass Calculated molar mass would be the same as the actual molar mass Suppose the freezing point of the solution w as determined to be 3.5 degreee C higher than it actually was. What would your calculated value for the molar mass of the solute be in this case? Show your work. What is the percent error between the molar mass from of this question and your experimental value you found in the experiment? Show your work. All of our unknowns were nonelectrolytes, so i = 1. What would the molar mass of your unknown have come out to if its value of i = 1.25? Show your work.Explanation / Answer
1. Freezing point experiment
dTf = i.Kf.(g/molar mass x kg solvent)
a. If the freezing point for t-BuOH is too low : calcualated molar mass would be larger than actual molar mass
b. If freezing point of mixture was too high : calculated molar mass would be smaller than actual molar mass
c. If water has contaminated t-BuOH : calculated molar mass would be larger than actual molar mass
2. For a solution with freezing point 3.5 oC higher than actual value
a. The calculated molar mass of solute would be lower than actual molar mass.
Freezing point dTf is inversely proportional to molar mass by equation,
dTf = i.Kf.(grams of solute/molar mass x kg of solvent)
b. The percent error would be = [(actual molar mass - calculated molar mass)/actual molar mass] x 100
3. With formula,
dTf = i.Kf.m
with i = 1, we have molar mass of solute
when i = 1.25.
The molar mass would be 1.25 times greater than actual value when i = 1.
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