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Chapter 11: Problem 64

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 mole fraction composition of the vapor at \(25^{\circ} C\) is: \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): 0.875, and \(\mathrm{CH}_{2} \mathrm{Br}_{2}\): 0.125.

Step by step solution

01

Calculate the mole fraction of each component in the liquid phase

We are given that we have 0.0300 moles of CH2Cl2 and 0.0500 moles of CH2Br2. Let's first determine the mole fraction of each component in the solution: Mole fraction of CH2Cl2, X_CH2Cl2 = moles of CH2Cl2 / (moles of CH2Cl2 + moles of CH2Br2) X_CH2Cl2 = 0.0300 / (0.0300 + 0.0500) = 0.375 Mole fraction of CH2Br2, X_CH2Br2 = moles of CH2Br2 / (moles of CH2Cl2 + moles of CH2Br2) X_CH2Br2 = 0.0500 / (0.0300 + 0.0500) = 0.625
02

Use Raoult's law to find the partial pressures of each component

Raoult's law states that the partial pressure of a component in the vapor phase is equal to the mole fraction of the component in the liquid phase multiplied by the vapor pressure of the pure component. We are given the vapor pressures of pure CH2Cl2 and pure CH2Br2. So we can calculate the partial pressures of each component in the vapor phase: Partial pressure of CH2Cl2 (P_CH2Cl2) = X_CH2Cl2 * Vapor pressure of pure CH2Cl2 P_CH2Cl2 = 0.375 * 133 torr = 49.88 torr Partial pressure of CH2Br2 (P_CH2Br2) = X_CH2Br2 * Vapor pressure of pure CH2Br2 P_CH2Br2 = 0.625 * 11.4 torr = 7.125 torr
03

Calculate the mole fractions of components in the vapor phase

The mole fraction of a component in the vapor is given by the partial pressure of the component divided by the total pressure of the vapor. We can find the mole fractions of each component in the vapor phase: Total pressure (P_total) = P_CH2Cl2 + P_CH2Br2 = 49.88 torr + 7.125 torr = 57.005 torr Mole fraction of CH2Cl2 in the vapor (Y_CH2Cl2) = P_CH2Cl2 / P_total Y_CH2Cl2 = 49.88 torr / 57.005 torr = 0.875 Mole fraction of CH2Br2 in the vapor (Y_CH2Br2) = P_CH2Br2 / P_total Y_CH2Br2 = 7.125 torr / 57.005 torr = 0.125 So, the mole fraction composition of the vapor at 25°C is: CH2Cl2: 0.875, and CH2Br2: 0.125.

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Most popular questions from this chapter

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).

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is $2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\( for camphor is \)40 .^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}$ ). Calculate the molality of the solution and the molar mass of reserpine.

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C} )\) b. a solution of glucose in water with $\chi_{\mathrm{CH}_{2} \mathrm{O}_{6}}=0.01$ c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}, 0 \mathrm{H}}=0.2\) (Consider the vapor pressure of both methanol \([143 \text { torr at }\) \(25^{\circ} \mathrm{C} ]\) and water.

A forensic chemist is given a white solid that is suspected of being pure cocaine $\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}, \text { molar mass }=303.35 \mathrm{g} / \mathrm{mol}\right)\( She dissolves \)1.22 \pm 0.01 \mathrm{g}\( of the solid in \)15.60 \pm 0.01 \mathrm{g}$ benzene. The freezing point is lowered by \(1.32 \pm 0.04^{\circ} \mathrm{C}\) a. What is the molar mass of the substance? Assuming that the percent uncertainty in the calculated molar mass is the same as the percent uncertainty in the temperature change, calculate the uncertainty in the molar mass. b. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine $\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}, \text { molar }\right.\( mass \)=299.36 \mathrm{g} / \mathrm{mol}$ )? c. Assuming that the absolute uncertainties in the measurements of temperature and mass remain unchanged, how could the chemist improve the precision of her results?

Explain the terms isotonic solution, crenation, and hemolysis.
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