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Chapter 11: Problem 58

A solution is prepared by mixing \(0.0300\) mole of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(0.0500\) mole of \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) 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 phase at 25°C is 0.875 mole fraction of \(CH_2Cl_2\) and 0.125 mole fraction of \(CH_2Br_2\).

Step by step solution

01

Calculate the total amount of moles of CH2Cl2 and CH2Br2

We know the moles of both substances in the solution and can sum them up to find the total moles: Total moles = moles of CH2Cl2 + moles of CH2Br2 Total moles = 0.0300 + 0.0500 Total moles = 0.0800
02

Calculate the mole fractions of CH2Cl2 and CH2Br2 in the liquid phase

We can now calculate the mole fractions of CH2Cl2 and CH2Br2 in the liquid phase using the given moles and total moles calculated in step 1: Mole fraction of CH2Cl2 (X1) = moles of CH2Cl2 / Total moles X1 = 0.0300 / 0.0800 X1 = 0.375 Mole fraction of CH2Br2 (X2) = moles of CH2Br2 / Total moles X2 = 0.0500 / 0.0800 X2 = 0.625
03

Apply Raoult's Law to find the partial pressures of CH2Cl2 and CH2Br2

Raoult's Law relates the partial pressures of the components in the vapor phase (P1 and P2) to their mole fractions in the liquid phase (X1 and X2) and their respective pure vapor pressures (P1° and P2°): P1 = X1 * P1° P2 = X2 * P2° Given the vapor pressures of pure CH2Cl2 (P1°) and CH2Br2 (P2°) as 133 and 11.4 torr, respectively, we can find the partial pressures for both substances in the vapor phase using their calculated mole fractions: P1 = 0.375 * 133 P1 = 49.875 torr P2 = 0.625 * 11.4 P2 = 7.125 torr
04

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

First, we need to find the total pressure in the vapor phase by adding the partial pressures of CH2Cl2 and CH2Br2: Total pressure = P1 + P2 Total pressure = 49.875 + 7.125 Total pressure = 57 torr Now, we can find the mole fractions of CH2Cl2 (Y1) and CH2Br2 (Y2) in the vapor phase using their respective partial pressures and the total pressure in the vapor phase: Mole fraction of CH2Cl2 in vapor phase (Y1) = P1 / Total pressure Y1 = 49.875 / 57 Y1 = 0.875 Mole fraction of CH2Br2 in vapor phase (Y2) = P2 / Total pressure Y2 = 7.125 / 57 Y2 = 0.125 The composition of the vapor phase at 25°C is 0.875 mole fraction of CH2Cl2 and 0.125 mole fraction of CH2Br2.

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