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Chapter 11: Problem 59

What is the composition of a methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) -propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) solution that has a vapor pressure of 174 torr at \(40^{\circ} \mathrm{C} ?\) At \(40^{\circ} \mathrm{C}\), the vapor pressures of pure methanol and pure propanol are 303 and \(44.6\) torr, respectively. Assume the solution is ideal.

Short Answer

Expert verified
The composition of the methanol-propanol solution at 40°C and 174 torr vapor pressure is approximately \(49.2\%\) methanol and \(56.2\%\) propanol.

Step by step solution

01

Understand Raoult's Law for ideal solutions.

Raoult's Law states that the partial pressure of a component in an ideal solution is equal to the product of its mole fraction and its constant vapor pressure. Mathematically, it can be written as: \[P_{A} = X_{A} \cdot P^{0}_{A}\] \[P_{B} = X_{B} \cdot P^{0}_{B}\] where \(P_{A}\) and \(P_{B}\) are the partial pressures of components A and B, \(X_{A}\) and \(X_{B}\) are their mole fractions in the solution, and \(P^{0}_{A}\) and \(P^{0}_{B}\) are their vapor pressures in pure state.
02

Set up the equations using given information.

In this problem, component A is methanol, and component B is propanol. We are given the following information: - The vapor pressure of the solution (174 torr) - The vapor pressure of pure methanol at 40°C (303 torr) - The vapor pressure of pure propanol at 40°C (44.6 torr) We can set up the following equations using Raoult's Law: \[P_{A} = X_{A} \cdot 303\] \[P_{B} = X_{B} \cdot 44.6\] Since the solution is ideal, the total vapor pressure is the sum of the partial pressures of each component: \[P_{total} = P_{A} + P_{B} = 174\]
03

Calculate the mole fractions.

We need to find the mole fractions of methanol (\(X_{A}\)) and propanol (\(X_{B}\)). We can use the mole fraction relationship \(X_{A} + X_{B} = 1\).
04

Solve for the unknowns.

First, express the mole fractions in terms of the partial pressures: \[X_{A} = \frac{P_{A}}{303}\] \[X_{B} = \frac{P_{B}}{44.6}\] Next, substitute these expressions in the mole fraction relationship: \[\frac{P_{A}}{303} + \frac{P_{B}}{44.6} = 1\] Now we can use the total vapor pressure relationship to relate the partial pressures: \[P_{A} = 174 - P_{B}\] Substitute this expression for \(P_{A}\) into our mole fraction equation: \[\frac{174 - P_{B}}{303} + \frac{P_{B}}{44.6} = 1\] Solve for \(P_{B}\) in the equation above: \[P_{B} \approx 25.05 \text{ torr}\] Now that we have \(P_B\), we can calculate \(P_A\): \[P_{A} = 174 - P_{B} \approx 148.95 \text{ torr}\]
05

Calculate the mole fractions of methanol and propanol.

Finally, we can calculate the mole fractions of methanol and propanol using the partial pressures calculated in Step 4: \[X_{A} = \frac{P_{A}}{303} \approx \frac{148.95}{303} \approx 0.492\] \[X_{B} = \frac{P_{B}}{44.6} \approx \frac{25.05}{44.6} \approx 0.562\] So, the composition of the methanol-propanol solution is approximately 49.2% methanol and 56.2% propanol at 40°C and 174 torr vapor pressure.

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Most popular questions from this chapter

A solution of phosphoric acid was made by dissolving \(10.0 \mathrm{~g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in \(100.0 \mathrm{~mL}\) water. The resulting volume was \(104 \mathrm{~mL}\). Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\).

Consider an aqueous solution containing sodium chloride that has a density of \(1.01 \mathrm{~g} / \mathrm{mL}\). Assume the solution behaves ideally. The freezing point of this solution at \(1.0 \mathrm{~atm}\) is \(-1.28^{\circ} \mathrm{C}\). Calculate the percent composition of this solution (by mass).

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