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Answer :
Using Raoult's Law, we can calculate that 3846.43 grams of C₂H₆O₂ are in the solution.
According to Raoult's law, the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.
[tex]P_{total} = X_A \times P_A + X_B \times P_B[/tex]
where:
[tex]P_{total}[/tex] is the total vapor pressure of the solution = 14.12 mm Hg,
[tex]X_A[/tex] and [tex]X_B[/tex] are the mole fractions of components A and B (in this case, water and C₂H₆O₂, respectively), and
[tex]P_A[/tex] (19.83 mm Hg) and [tex]P_B[/tex] are the vapor pressures of components A and B in their pure states.
Let the unknown quantity of C₂H₆O₂ in the solution be "x" grams.
Calculating the mole fraction of C₂H₆O₂ ([tex]X_B[/tex]) in the solution
The mole fraction of C₂H₆O₂ ([tex]X_B[/tex]) can be calculated from the following formula:
[tex]X_B[/tex] = moles of C₂H₆O₂ / total moles in the solution
Since we are given that the solution contains 100 grams of water and "x" grams of C₂H₆O₂, the total moles in the solution can be calculated as follows:
Total moles = moles of water + moles of C₂H₆O₂
Moles of water = 100 grams / molar mass of water (18.015 g/mol)
Moles of C₂H₆O₂ = x / molar mass of C₂H₆O₂ (46.07 g/mol)
Further, rearrange and use Raoult's law to calculate "x":
[tex]X_B[/tex] = ([tex]P_{total}[/tex] - [tex]X_A[/tex] * [tex]P_A[/tex]) / [tex]P_A[/tex]
[tex]X_A[/tex] = moles of water / total moles in the solution
[tex]X_B[/tex] = ([tex]P_{total}[/tex] - [tex]X_A[/tex] * [tex]P_A[/tex]) / [tex]P_B[/tex]
Moles of water = 100 g / 18.015 g/mol ≈ 5.547 mol
Moles of Câ‚‚H₆Oâ‚‚ = x g / 46.07 g/mol ≈ x / 46.07 mol
Total moles in the solution = 5.547 mol + (x / 46.07) mol
[tex]X_A[/tex] = [tex]\frac{5.547 mol }{5.547 mol + \frac{x}{46.07 mol} }[/tex]
[tex]X_A[/tex] = 5.547 / (5.547 + 0.0215x)
[tex]X_B[/tex] = (14.12 mm Hg - [tex]X_A[/tex] * 19.83 mm Hg) / 19.83 mm Hg
Now, solve for "x":
[tex]X_B[/tex] = (14.12 - 19.83 × 5.547 / (5.547 + 0.0215x)) / 19.83
[tex]X_B[/tex] ≈ -102.56781 / (19.83 + 0.0215x)
We know that [tex]X_B[/tex] cannot be negative.
[tex]X_B[/tex] ≈ 102.56781 / (19.83 + 0.0215x)
102.56781 / (19.83 + 0.0215x) = 1 (since it is a mole fraction)
0.0215x = 82.73781
x = 82.73781 / 0.0215
x ≈ 3846.428 g
Therefore, approximately 3846.43 grams of C₂H₆O₂ are in the solution.
To learn more about the molar fraction
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