A solution is prepared by mixing 1.000 mole of methanol $\left(\mathrm{CH}_{3} \mathrm{OH}\right)\( and 3.18 moles of propanol \)\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)$ What is the composition of the vapor (in mole fractions) at $40^{\circ} \mathrm{C} ? \mathrm{At} 40^{\circ} \mathrm{C},$ the vapor pressure of pure methanol is 303 torr, and the vapor pressure of pure propanol is 44.6 torr.

Short Answer

Expert verified
The mole fractions of methanol and propanol in the solution are \(x_{CH_3OH} = \frac{1.000}{4.18}\) and \(x_{CH_3CH_2CH_2OH} = \frac{3.18}{4.18}\), respectively. After calculating the partial pressures using Raoult's Law, and the total vapor pressure, the mole fractions of methanol and propanol in the vapor phase can be found by dividing their respective partial pressures by the total vapor pressure. The mole fractions of methanol and propanol in the vapor phase represent the composition of the vapor at 40°C.

Step by step solution

01

Understand Raoult's Law

Raoult's Law states that the partial pressure of a component in a solution is equal to its mole fraction multiplied by its vapor pressure in the pure state at the same temperature. Mathematically, it can be represented as: \(P_i = x_i P_i^*\) Where: - \(P_i\) is the partial pressure of component i in the solution - \(x_i\) is the mole fraction of component i in the solution - \(P_i^*\) is the vapor pressure of pure component i at the same temperature
02

Calculate the mole fraction of each component in the solution

We are given 1.000 mole of methanol and 3.18 moles of propanol are mixed together. First, let's find the mole fractions of each component in the solution. Total moles = moles of methanol + moles of propanol Total moles = 1.000 + 3.18 = 4.18 Mole fraction of methanol, \(x_{CH_3OH}\) = \(\frac{1.000}{4.18}\) Mole fraction of propanol, \(x_{CH_3CH_2CH_2OH}\) = \(\frac{3.18}{4.18}\)
03

Calculate the partial pressure of each component in the solution

Using the mole fractions of methanol and propanol, we can calculate their partial pressures using Raoult's Law: For methanol: \(P_{CH_3OH} = x_{CH_3OH} P_{CH_3OH}^*\) \(P_{CH_3OH} = (\frac{1.000}{4.18})(303\, \mathrm{torr})\) For propanol: \(P_{CH_3CH_2CH_2OH} = x_{CH_3CH_2CH_2OH} P_{CH_3CH_2CH_2OH}^*\) \(P_{CH_3CH_2CH_2OH} = (\frac{3.18}{4.18})(44.6\, \mathrm{torr})\)
04

Calculate the total vapor pressure and mole fractions in the vapor phase

Now, let's calculate the total vapor pressure of the solution by adding the partial pressures of methanol and propanol: Total vapor pressure = \(P_{CH_3OH} + P_{CH_3CH_2CH_2OH}\) Finally, we can calculate mole fractions in the vapor phase: Mole fraction of methanol in the vapor phase = \(\frac{P_{CH_3OH}}{Total\, vapor\, pressure}\) Mole fraction of propanol in the vapor phase = \(\frac{P_{CH_3CH_2CH_2OH}}{Total\, vapor\, pressure}\) These mole fractions represent the composition of the vapor at 40°C. Make sure to calculate each value using the expressions derived above to get the final numerical results.

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