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Chapter 11: Problem 58

A solution is prepared by mixing \(0.0300 \mathrm{~mol} \mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(0.0500\) \(\mathrm{mol} \mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at \(25^{\circ} \mathrm{C}\). At \(25^{\circ} \mathrm{C}\), the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and \(11.4\) torr, respectively.

Short Answer

Expert verified
The composition of the vapor at 25°C is: Mole fraction of CH₂Cl₂ (Y₁) = 0.875 Mole fraction of CH₂Br₂ (Y₂) = 0.125

Step by step solution

01

Calculate the mole fractions of CH₂Cl₂ and CH₂Br₂ in the solution.

First, we need to find the mole fractions of CH₂Cl₂ and CH₂Br₂ in the solution. To do this, we divide the moles of each component by the total moles in the solution. Total moles = moles of CH₂Cl₂ + moles of CH₂Br₂ Total moles = 0.0300 mol + 0.0500 mol = 0.0800 mol Now, we find the mole fractions: Mole fraction of CH₂Cl₂ (X₁) = (moles of CH₂Cl₂) / (total moles) = 0.0300 mol / 0.0800 mol = 0.375 Mole fraction of CH₂Br₂ (X₂) = (moles of CH₂Br₂) / (total moles) = 0.0500 mol / 0.0800 mol = 0.625
02

Calculate the partial pressures of CH₂Cl₂ and CH₂Br₂.

Next, we will use Raoult's Law to calculate the partial pressures (P₁ and P₂) of CH₂Cl₂ and CH₂Br₂ in the vapor phase. Raoult's Law states: P_i = X_i * P_i° where P_i is the partial pressure of component i, X_i is the mole fraction of component i in the solution, and P_i° is the vapor pressure of the pure component i. For CH₂Cl₂ (component 1): P₁ = X₁ * P₁° = 0.375 * 133 torr = 49.875 torr For CH₂Br₂ (component 2): P₂ = X₂ * P₂° = 0.625 * 11.4 torr = 7.125 torr
03

Calculate the total pressure of the vapor phase.

Now we will find the total pressure (P_total) of the vapor phase by adding the partial pressures of each component: P_total = P₁ + P₂ = 49.875 torr + 7.125 torr = 57.000 torr
04

Calculate the mole fractions of CH₂Cl₂ and CH₂Br₂ in the vapor phase.

Finally, we will use the partial pressures to calculate the mole fractions of each component in the vapor phase: Mole fraction of CH₂Cl₂ in vapor (Y₁) = P₁ / P_total = 49.875 torr / 57.000 torr = 0.875 Mole fraction of CH₂Br₂ in vapor (Y₂) = P₂ / P_total = 7.125 torr / 57.000 torr = 0.125 The composition of the vapor at 25°C is: Mole fraction of CH₂Cl₂ (Y₁) = 0.875 Mole fraction of CH₂Br₂ (Y₂) = 0.125

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Most popular questions from this chapter

From the following: pure water solution of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(m=0.01)\) in water solution of \(\mathrm{NaCl}(m=0.01)\) in water solution of \(\mathrm{CaCl}_{2}(m=0.01)\) in water choose the one with the a. highest freezing point. d. lowest boiling point. b. lowest freezing point. e. highest osmotic pressure. c. highest boiling point.

What volume of a \(0.580 M\) solution of \(\mathrm{CaCl}_{2}\) contains \(1.28 \mathrm{~g}\) solute?

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A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\). The freezing point of the solution is \(2.81^{\circ} \mathrm{C}\). What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\).
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