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Chapter 10: Problem 55

A solution is made by dissolving \(25.8 \mathrm{g}\) urea \(\left(\mathrm{CH}_{4} \mathrm{N}_{2} \mathrm{O}\right),\) a nonelectrolyte, in \(275 \mathrm{g}\) water. Calculate the vapor pressures of this solution at \(25^{\circ} \mathrm{C}\) and \(45^{\circ} \mathrm{C}\). (The vapor pressure of pure water is \(\left.23.8 \text { torr at } 25^{\circ} \mathrm{C} \text { and } 71.9 \text { torr at } 45^{\circ} \mathrm{C} .\right)\)

Short Answer

Expert verified
The vapor pressures of the urea-water solution at the given temperatures are: At 25°C, the vapor pressure is 23.1 torr, and at 45°C, the vapor pressure is 69.9 torr.

Step by step solution

01

(Step 1: Find the moles of urea and water)

First, we need to calculate the moles of urea and water in the solution. Calculate the moles of urea by dividing the mass of urea by its molar mass, and the moles of water by dividing the mass of water by the molar mass of water. Molar mass of urea: 12 (for C) + (4 x 1) (for H) + (2 x 14) (for N) + 16 (for O) = 60 g/mol Moles of urea: \(\frac{25.8\, g}{60\, g/mol} = 0.430\, mol\) Molar mass of water: (2 x 1) (for H) + 16 (for O) = 18 g/mol Moles of water: \(\frac{275\, g}{18\, g/mol} = 15.3\, mol\)
02

(Step 2: Calculate the mole fraction of urea and water)

Next, we need to determine the mole fraction of urea and water in the solution. The mole fraction is the ratio of the number of moles of a particular component to the total number of moles. Mole fraction of urea (\(X_{urea}\)) = \(\frac{moles\, urea}{moles\, urea + moles\, water} = \frac{0.430}{0.430 + 15.3} = 0.0273\) Mole fraction of water (\(X_{H_2O}\)) = \(\frac{moles\, water}{moles\, urea + moles\, water} = \frac{15.3}{0.430 + 15.3} = 0.9727\)
03

(Step 3: Apply Raoult's law)

Now, we can use Raoult's law to calculate the vapor pressures of the solution at 25°C and 45°C. According to Raoult's law, the partial pressure of a component in a non-ideal solution is equal to the product of its mole fraction and the vapor pressure of the pure component. At 25°C: Vapor pressure of the solution = \(X_{H_2O} * P^*_H{_2}O\) where \(P^*_{H_2O}\) is the vapor pressure of pure water at 25°C. Vapor pressure of the solution = \(0.9727 * 23.8\, torr = 23.1\, torr\) At 45°C: Vapor pressure of the solution = \(X_{H_2O} * P^*_H{_2}O\) where \(P^*_{H_2O}\) is the vapor pressure of pure water at 45°C. Vapor pressure of the solution = \(0.9727 * 71.9\, torr = 69.9\, torr\)
04

(Step 4: Report the vapor pressures)

Finally, we can report the vapor pressures of the urea-water solution at the given temperature: At 25°C: The vapor pressure of the solution is 23.1 torr. At 45°C: The vapor pressure of the solution is 69.9 torr.

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Most popular questions from this chapter

An aqueous solution of 10.00 g of catalase, an enzyme found in the liver, has a volume of \(1.00 \mathrm{L}\) at \(27^{\circ} \mathrm{C}\). The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be 0.745 torr. Calculate the molar mass of catalase.

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