Question

A chemist dissolves \(300 \mathrm{~g}\) of urea in \(1000 \mathrm{~g}\) of water. Urea is \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). Assuming the solution obeys Raoult's'law, determine the following: a) The vapor pressure of the solvent at \(0^{\circ}\) and \(100^{\circ} \mathrm{C}\) and b) The boiling and freezing point of the solution. The vapor pressure of pure water is \(4.6 \mathrm{~mm}\) and \(760 \mathrm{~mm}\) at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively and the \(\mathrm{K}_{\mathrm{f}}=(1.86\) \(\mathrm{C} / \mathrm{mole}\) ) and \(\mathrm{K}_{\mathrm{b}}=(52 \mathrm{C} / \mathrm{mole})\)

Short answer

The vapor pressure of the solvent at \(0^{\circ}\) and \(100^{\circ} \mathrm{C}\) is approximately \(4.6 \mathrm{~mm}\) and \(760 \mathrm{~mm}\), respectively. The boiling point of the solution is \(360^{\circ} \mathrm{C}\), and the freezing point is \(-9.3^{\circ} \mathrm{C}\).

Step-by-step solution

Calculate the molality of the solution

We are given the mass of urea and water, so we need to determine the moles of each and then calculate the molality. Urea has a molecular formula of \(NH_{2}CONH_{2}\) and a molar mass of \(12 + 2(1) + 14 + 16 + 14 + 2(1) = 60 \mathrm{~g/mol}\). Thus, the number of moles of urea in the solution can be calculated as: \[\text{moles of urea} = \frac{300 \mathrm{~g}}{60 \mathrm{~g/mol}} = 5\mathrm{~moles}\] Now, we can calculate the molality of the solution: \[\text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{5\mathrm{~moles}}{1\mathrm{~kg}} = 5\mathrm{~mol/kg}\]

Use Raoult's Law to determine the vapor pressure of the solvent

Raoult's Law states that the vapor pressure of the solvent in a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. The mole fraction is defined as the ratio of moles of solvent to the total moles of both solute and solvent. Since the amount of water is so much greater than that of urea, we can assume that the mole fraction of water is approximately equal to 1. Thus, we can calculate the vapor pressure of the solvent at 0°C and 100°C: At 0°C: \[P_{solvent} = P_{pure\ water} * X_{water} \approx 4.6\mathrm{~mm} * 1 = 4.6\mathrm{~mm}\] At 100°C: \[P_{solvent} = P_{pure\ water} * X_{water} \approx 760\mathrm{~mm} * 1 = 760\mathrm{~mm}\]

Determine the boiling and freezing points of the solution using depression constants

The boiling point elevation and freezing point depression can be calculated using the following formulas, where ΔTb and ΔTf are the changes in the boiling and freezing points, Kb and Kf are the boiling and freezing point depression constants, and molality is the molality of the solution: ΔTb = Kb * molality ΔTf = Kf * molality Given the values of Kb and Kf, we can now calculate the boiling point elevation and freezing point depression as follows: ΔTb = 52°C/mol * 5 mol/kg = 260°C ΔTf = 1.86°C/mol * 5 mol/kg = 9.3°C Now, we can add ΔTb to the boiling point and subtract ΔTf from the freezing point of water to get the boiling and freezing points of the solution: Boiling point of the solution = 100°C + 260°C = 360°C Freezing point of the solution = 0°C - 9.3°C = -9.3°C

Key concepts

Molality Calculation

Molality is a crucial concentration measure in chemistry, especially when dealing with colligative properties such as boiling point elevation and freezing point depression. It is defined as the number of moles of solute per kilogram of solvent.

To calculate molality, you need two pieces of information: the mass of the solute being dissolved and the mass of the solvent used to dissolve the solute. Once you have this, the procedure is straightforward. You convert the solute's mass to moles by dividing by its molar mass. Then, you convert the solvent's mass from grams to kilograms. Finally, you divide the moles of solute by the kilograms of solvent. In our example, the molality of urea in water has been calculated as \[ \text{molality} = \frac{5 \text{~moles of urea}}{1 \text{~kg of water}} = 5 \text{~mol/kg} \]This calculation is crucial because molality does not change with temperature, making it a reliable metric for studying temperature-dependent properties.

Vapor Pressure Determination

Vapor pressure is an indicator of a liquid's evaporation rate and is influenced by the presence of a non-volatile solute. Raoult's Law allows us to determine the vapor pressure of a solvent in a solution

According to Raoult's Law, the vapor pressure of the solvent decreases proportionally to the presence of a solute. We assume that the pure solvent has a certain vapor pressure, and the solution's vapor pressure equals the product of the pure solvent's vapor pressure and the mole fraction of the solvent. The mole fraction, in turn, is the ratio of moles of the solvent to the total moles in solution.

For dilute solutions or when the solute is nonvolatile (like urea), the vapor pressure of the solvent is almost the same as that of the pure solvent. In our example, the vapor pressure of water in the solution at both 0°C and 100°C remains nearly unchanged:

Boiling Point Elevation

Boiling point elevation occurs when a non-volatile solute is dissolved in a solvent, raising the solvent's boiling point. This phenomenon is one of the colligative properties, meaning it depends on the quantity but not the identity of the solute particles.

The formula to calculate the boiling point elevation is \[ \Delta T_b = K_b \times \text{molality} \] where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the ebullioscopic constant (boiling point elevation constant), and molality is the concentration of the solution in terms of molality.

In our textbook example, the boiling point elevation for the urea solution is significant: \[ \Delta T_b = 52 \ degree C/mol \times 5 mol/kg = 260 \ degree C \]It's important to ensure that the boiling point elevation does not exceed the practical or theoretical limits of the solvent's boiling point.

Freezing Point Depression

Freezing point depression is the process by which the freezing point of a solvent is lowered by the addition of a solute. Like boiling point elevation, it's a colligative property, relevant in many practical applications, including the manufacturing of antifreeze.

To calculate the freezing point depression, we use the formula: \[ \Delta T_f = K_f \times \text{molality} \] where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant (freezing point depression constant), and molality is the molality of the solution.

For the given urea solution, the freezing point drops by: \[ \Delta T_f = 1.86 \ degree C/mol \times 5 mol/kg = 9.3 \ degree C \]This means that water, which normally freezes at 0°C, will now freeze at approximately -9.3°C. Freezing point depression can provide critical information for designing solutions to resist colder environments without solidifying.