Question

Calculate the vapor pressure (in torr) at 298 k in a solution prepared by dissolving 15.3 g of the non-volatile non-electrolye urea {co(nh2)2} in 107 g of water. the vapor pressure of water at 298 k is 23.76 torr. enter your answer to 2 decimal places

Answer 1

The vapor pressure of the solution at 298 K is 22.80 torr. This is calculated by determining the mole fraction of water in the solution using the given masses of urea and water, and applying Raoult's law with the vapor pressure of pure water.

Step-by-step explanation:

To calculate the vapor pressure of the solution at 298 K, we must apply Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. In this case, we are given the mass of urea (15.3 g) and the mass of water (107 g), along with the vapor pressure of pure water at 298 K, which is 23.76 torr.

First, we need to convert the masses to moles. The molar mass of water (H2O) is about 18.015 g/mol, and the molar mass of urea (CO(NH2)2) is approximately 60.06 g/mol. Therefore:

• Number of moles of water: 107 g / 18.015 g/mol = 5.942 moles
• Number of moles of urea: 15.3 g / 60.06 g/mol = 0.255 moles

The total number of moles in the solution is the sum of moles of water and urea, which is 5.942 + 0.255 = 6.197 moles. Now we determine the mole fraction of water:

Mole fraction of water (XH2O) = Moles of water / Total moles = 5.942 / 6.197 ≈ 0.959

Using Raoult's law, the vapor pressure of the solution (Psolution) is the product of the mole fraction of water and the vapor pressure of pure water:

Psolution = XH2O * Vapor pressure of pure water = 0.959 * 23.76 torr ≈ 22.80 torr

Therefore, the vapor pressure of the solution at 298 K is 22.80 torr.

Answer 2

In colligative properties like vapor pressure, boiling point, freezing point and osmotic pressure, only the quantity of solute may affect their values. Their identity does not affect these properties. When solute is added to a solution, the vapor pressure decreases, that's why it is called vapor pressure lowering. The equation is

ΔP = xP°, where ΔP is the difference of the vapor pressure of the solution and of the solvent. P° is the vapor pressure of the pure solvent while x is the mole fraction of solute. The molar mass of urea is 60.06 g/mol while that of the water is 18 g/mol.

x =mol urea/total mol = (15.3/60.06)/[(15.3/60.06)+(107/18)]
x = 0.0411

Then,
ΔP = xP°=0.0411*23.76 = 0.976 torr

Therefore, the vapor pressure of the solution is 23.76 - 0.976 = 22.78 Torr.