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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Fraction
The concept of mole fraction plays a vital role in solution chemistry. It is a way to express the concentration of a component in a mixture. To calculate the mole fraction, you divide the number of moles of the component by the total number of moles present in the solution. This ratio shows the proportion of each component within the mixture.

In the context of the given exercise involving urea and water, the mole fraction helps determine how the presence of urea affects the overall properties of the solution. For instance, you need to find the moles of urea and water separately using their respective molar masses:
  • For urea, with a molar mass of 60 g/mol, divide the mass of urea (25.8 g) by 60 g/mol to get 0.430 moles.
  • For water, with a molar mass of 18 g/mol, divide the mass of water (275 g) by 18 g/mol to get 15.3 moles.
Once you have the moles, you can easily calculate the mole fractions:
  • The mole fraction of urea \(X_{urea}\) is \( \frac{0.430 \, \text{mol}}{0.430 \, \text{mol} + 15.3 \, \text{mol}} = 0.0273 \).
  • The mole fraction of water \(X_{H_2O}\) is \( \frac{15.3 \, \text{mol}}{0.430 \, \text{mol} + 15.3 \, \text{mol}} = 0.9727 \).
Understanding the mole fraction helps in calculating other important properties like vapor pressure, as it reflects how much each component influences the behavior of the solution.
Vapor Pressure
Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid phase at a specific temperature. It is a crucial factor in understanding how substances behave in solutions. Raoult's Law is used to predict the vapor pressure of a solution, especially when dealing with nonelectrolyte solutions.

Raoult's Law states that the partial vapor pressure of a solvent in a solution is equal to the product of the mole fraction of the solvent and the vapor pressure of the pure solvent. This relationship helps us to find out how the addition of a solute like urea affects the vapor pressure of the solvent (water in this case).
  • At 25°C, using the mole fraction of water \(X_{H_2O} = 0.9727\) and the pure water vapor pressure of 23.8 torr, the solution's vapor pressure is \(0.9727 \times 23.8 \, \text{torr} = 23.1 \, \text{torr}\).
  • Similarly, at 45°C, with the pure water vapor pressure of 71.9 torr, the vapor pressure becomes \(0.9727 \times 71.9 \, \text{torr} = 69.9 \, \text{torr}\).
The vapor pressure of the solution is lower than the pure solvent because of the addition of a non-volatile solute like urea. This lowering of vapor pressure also results in an elevation of boiling point and depression of freezing point, properties which are often explored in solution chemistry.
Solution Chemistry
Solution chemistry examines how different substances interact when forming a homogenous mixture. It looks into how solutes dissolve in solvents, forming solutions, and how these solutions behave in various conditions.

In our exercise, urea acts as the solute and water as the solvent. Since urea is nonelectrolyte, it dissolves in water without breaking into ions. This results in a solution that still retains the inherent characteristics of the mixture, but with modified properties like vapor pressure.

The solution's properties, such as boiling point elevation or vapor pressure reduction, are colligative properties. This means they depend on the ratio of solute to solvent rather than the nature of the solute. By understanding the mole fraction, we use it to predict such changes accurately via Raoult's Law.
  • Non-volatile solutes reduce the vapor pressure of the solution compared to the pure solvent.
  • The extent of the change in properties hinges on the mole fraction of the solute.
Through the lens of solution chemistry, the effect of a solute on a solvent's physical properties can be predicted and calculated, helping in numerous practical applications ranging from industrial processes to everyday solutions.

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Most popular questions from this chapter

Assume that you place a freshwater plant into a saltwater solution and examine it under a microscope. What happens to the plant cells? What if you placed a saltwater plant in pure water? Explain. Draw pictures to illustrate your explanations.

In the winter of \(1994,\) record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of \(-29^{\circ} \mathrm{F}\) was registered. At this temperature can salting icy roads with \(\mathrm{CaCl}_{2}\) be effective in melting the ice? a. Assume \(i=3.00\) for \(\mathrm{CaCl}_{2}\) b. Assume the average value of \(i\) from Exercise 87 (The solubility of \(\mathrm{CaCl}_{2}\) in cold water is \(74.5 \mathrm{g}\) per \(100.0 \mathrm{g}\) water.

For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

A solid mixture contains \(\mathrm{MgCl}_{2}\) and \(\mathrm{NaCl}\). When \(0.5000 \mathrm{g}\) of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at \(25.0^{\circ} \mathrm{C}\) is observed to be 0.3950 atm. What is the mass percent of \(\mathrm{MgCl}_{2}\) in the solid? (Assume ideal behavior for the solution.)

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