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A solution has 2 moles of glucose (C6H12O6) in 3 kg of benzene (C6H6) at 20 C. What is the vapor pressure of the SOLUTION?

(Benzene vapor pressure at 20 C is 75 torr.)



Sagot :

Given:

moles of solute = 2 mol

mass of solvent = 3 kg = 3000 g

[tex]P_{\text{solvent}} = \text{75 torr}[/tex]

solute: glucose

solvent: benzene

Required:

[tex]P_{\text{solution}}[/tex]

Solution:

First Step: Calculate the molar mass of solvent.

molar mass of solvent = (12.01 g/mol × 6) + (1.008 g/mol × 6)

molar mass of solvent = 78.108 g/mol

Second Step: Calculate the number of moles of solvent.

[tex]\text{moles of solvent} = \dfrac{\text{mass of solvent}}{\text{molar mass of solvent}}[/tex]

[tex]\text{moles of solvent} = \dfrac{\text{3000 g}}{\text{78.108 g/mol}}[/tex]

moles of solvent = 38.40836 mol

Third Step: Calculate the mole fraction of solvent.

[tex]X_{\text{solvent}} = \dfrac{\text{moles of solvent}}{\text{moles of solute + moles of solvent}}[/tex]

[tex]X_{\text{solvent}} = \dfrac{\text{38.40836 mol}}{\text{2 mol + 38.40836 mol}}[/tex]

[tex]X_{\text{solvent}} = 0.9505053[/tex]

Last Step: Calculate the vapor pressure of the solution.

[tex]P_{\text{solution}} = X_{\text{solvent}}P_{\text{solvent}}[/tex]

[tex]P_{\text{solution}} = (0.9505053)(\text{75 torr})[/tex]

[tex]\boxed{P_{\text{solution}} = \text{71 torr}}[/tex]

Answer:

[tex]P_{\text{solution}} = \text{71 torr}[/tex]

[tex]\\[/tex]

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