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Trial #1 Trial #2 Trial #3 Total Volume of solution at equilibrium 11.9 mL 12.9 mL...

Trial #1 Trial #2 Trial #3
Total Volume of solution at equilibrium 11.9 mL 12.9 mL 12.2 mL
Volume of water added to reach equilibrium 8.9 mL 9.7 mL 9.2 mL
Mass of Urea 3.0570g 3.3634g 3.1331g

Using the formula K = [urea] / Xsolvent:

Calculate the mole fraction of water for each trial and calculate the new equilibrium constant based on the mole fraction.

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Homework Answers

Answer #1

Trial 1

moles of H2O =

=0.494 mol

Moles of urea =

=0.051 mol

mole fraction of water =

=0.906

Equilibrium constant =

[urea] = moles/volume of solution = 0.051/0.0119 = 4.29 M

----------------------------------------

Trial 2

moles of H2O =

=0.539 mol

Moles of urea =

=0.056 mol

mole fraction of water =

=0.906

Equilibrium constant =

[urea] = moles/volume of solution = 0.056/0.0129 = 4.34M

--------------------------------------------------------------

Trial3

moles of H2O =

=0.511 mol

Moles of urea =

=0.0522 mol

mole fraction of water =

=0.907

Equilibrium constant =

[urea] = moles/volume of solution = 0.0522/0.0122 = 4.28 M

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