Question

Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation of the solute. Express your answer in atmospheres to three significant figures. Pvap = atm

Answer 1

The question lacks the necessary information, such as mole fraction or solute concentration, to calculate the vapor pressure above a solution using Raoult's law. The general form of Raoult's law is given for reference.

Step-by-step explanation:

To calculate the vapor pressure of water above a solution, one would typically use Raoult's law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. However, the question does not provide enough details such as the concentration of the solute or the mole fraction of the solvent. Therefore, we can only provide the general form of the equation:

Psolution = Ppure solvent * Xsolvent

Where Psolution is the vapor pressure of the solvent above the solution, Ppure solvent is the vapor pressure of the pure solvent, and Xsolvent is the mole fraction of the solvent in the solution. To solve the question completely, one would need the mole fraction or the amount of solute and solvent used.

Answer 2

The question is incomplete, the solute was not given.

Let the solute be K₂CrO₄ and the solvent be water

Complete Question should be like this:

The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.

Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.

Pvap = ________atm

Answer:

Pvap (of water above the solution) = 0.0306 atm

Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

Step-by-step explanation:

Given

volume of solution = 1 Litre = 1000 mL of the solution

density of the solution = 1.063 g/mL

concentration of the solution= 0.438M

temperature of the solution= 298 K

vapour pressure of pure water = 0.0313atm

Recall: density = mass/volume

∴mass of solution = volume x density

m = 1000 x 1.063 = 1063 g

To calculate the moles of K₂CrO₄ = volume x concentration

= 1 x 0.438 = 0.438 mol

Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g

Mass of water = mass of solution - mass of K₂CrO₄

= 1063 - 85.055 = 977.945 g

moles of water = mass/molar mass

∴ moles of water = 977.945/18.02 = 54.27 mol

Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution

moles of ions = 3 x moles of K₂CrO₄

= 3 x 0.438 = 1.314 mol

Vapor pressure of solution = mole fraction of water x vapor pressure of water

= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm