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Chapter 13: Problem 68

At \(20^{\circ} \mathrm{C},\) the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that ben- benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of 35 torr at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Short Answer

Expert verified
In conclusion, the composition of the solution in mole fractions is: \(x_{C_6H_6} = 0.467\) for benzene and \(x_{C_7H_8} = 0\) for toluene. The mole fraction of benzene in the vapor above the solution is also 0.467.

Step by step solution

01

Recall Raoult's Law for mole fraction

For an ideal solution, Raoult's Law states that the partial pressure of a component in a mixture is equal to the mole fraction of that component multiplied by its vapor pressure as a pure substance: \(P_A = x_A P_A^*\) Where: - \(P_A\) is the partial pressure of component A in the mixture - \(x_A\) is the mole fraction of component A in the mixture - \(P_A^*\) is the vapor pressure of component A as a pure substance
02

Calculate the mole fraction of benzene in solution using Raoult's Law

Given that the vapor pressure of benzene (\(C_6H_6\)) is 75 torr, and the vapor pressure of toluene (\(C_7H_8\)) is 22 torr, we can write down the equations: - For benzene: \(P_{C_6H_6} = x_{C_6H_6} \times 75\) - For toluene: \(P_{C_7H_8} = x_{C_7H_8} \times 22\) - Total pressure: \(P_{total} = P_{C_6H_6} + P_{C_7H_8} = 35\) Since the sum of the mole fractions is equal to 1, we can write: \(x_{C_6H_6} + x_{C_7H_8} = 1\) \(x_{C_7H_8} = 1 - x_{C_6H_6}\) Substitute the equation for toluene's mole fraction into the total pressure equation: \(P_{total} = 75 x_{C_6H_6} + 22 (1 - x_{C_6H_6})\) Now solve for the mole fraction of benzene (\(x_{C_6H_6}\)).
03

Solve for the mole fraction of benzene in solution

Now we can solve the equation for \(x_{C_6H_6}\): \begin{align*} 35 &= 75x_{C_6H_6} + 22 (1 - x_{C_6H_6}) \\ 35 &= 75x_{C_6H_6}+ 22- 22x_{C_6H_6} \\ 53 &= 53x_{C_6H_6} \\ x_{C_6H_6} &= 1 \end{align*} The mole fraction of benzene in the solution is 1 and the mole fraction of toluene is 0, which means it is a pure benzene solution.
04

Calculate the mole fraction of benzene in the vapor above the solution using Raoult's Law

From Raoult's Law: \(P_{C_6H_6} = x_{C_6H_6} \times 75\) Since the solution is pure benzene, we can now solve for the mole fraction in the vapor: \begin{align*} P_{C_6H_6} &= x_{C_6H_6} \times 75 \\ \frac{P_{C_6H_6}}{75} &= x_{C_6H_6} \end{align*} Plugging in the values we have: \begin{align*} x_{C_6H_6} &= \frac{35}{75} \\ x_{C_6H_6} &= 0.467 \end{align*} Since the vapor phase consists of just benzene, its mole fraction is equal to that in the solution which is 0.467. In conclusion, the composition of the solution in mole fractions is: \(x_{C_6H_6} = 0.467\) for benzene and \(x_{C_7H_8} = 0\) for toluene. The mole fraction of benzene in the vapor above the solution is also 0.467.

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Most popular questions from this chapter

A dilute aqueous solution of an organic compound soluble in water is formed by dissolving 2.35 g of the compound in water to form 0.250 L of solution. The resulting solution has an osmotic pressure of 0.605 atm at \(25^{\circ} \mathrm{C}\) . Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

Indicate the principal type of solute-solvent interaction in each of the following solutions and rank the solutions from weakest to strongest solute- solvent interaction: (a) KCl in water, (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),\) (c) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water.

The presence of the radioactive gas radon (Rn) in well water presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at \(30^{\circ} \mathrm{C}\) is \(7.27 \times 10^{-3} \mathrm{M},\) what is the Henry's law constant for radon in water at this temperature? (b) A sample consisting of various gases contains \(3.5 \times 10^{-6}\) mole fraction of radon. This gas at a total pressure of 32 atm is shaken with water at \(30^{\circ} \mathrm{C} .\) Calculate the molar concentration of radon in the water.

The density of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(0.867 \mathrm{g} / \mathrm{mL},\) and the density of thiophene \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{S}\right)\) is 1.065 \(\mathrm{g} / \mathrm{mL}\) . A solution is made by dissolving 8.10 \(\mathrm{g}\) of thiophene in 250.0 \(\mathrm{mL}\) of toluene.(a) Calculate the molefraction of thiophene in the solution. (b) Calculate the molality of thiophene in the solution. (c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?

Choose the best answer: A colloidal dispersion of one liquid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol.
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