Question

Liquid A and liquid B form a solution that behaves ideally according to Raoult's law. The vapor pressures of the pure substances A and B are 218 torr and 135 torr, respectively. Determine the vapor pressure over the solution if 1.28 moles of liquid A is added to 5.30 moles of liquid B. 1. 42.4 torr 2. 151 torr 3. 188 torr 4. 202 torr 5. 760 torr

Answer 1

Answer:

Vapor pressure of solution → 151.1 Torr

Option 2.

Explanation:

Raoult's Law is relationed to colligative property about vapor pressure. A determined solute, can make, the vapor pressure of solution decreases.

ΔP = P° . Xm

where Xm is the mole fraction of solute, P° (vapor pressure of pure solvent)

and ΔP = Vapor pressure of pure solvent - Vapor pressure of solution.

In order to determine the vapor pressure of solution, we need to determine, the vapor pressure of B and A in the solution

B's pressure = P° B . Xm

When we add A to B, A works as the solute and B, as the solvent.

Vapor pressure of pure B is 135 torr. (P° B)

In order to determine, the Xm, we use the moles of A and B

Xm = 5.3 mol of B / (1.28 + 5.3) → 0.806

B's pressure = 135 Torr . 0.806 → 108.81 Torr

If mole fraction of B is 0.806, mole fraction for A (solute) will be (1 - 0.806)

A's pressure = 218 Torr . 0.194 → 42.3 Torr

Vapor pressure of solution is sum of vapor pressures of solute + solvent.

Vapor pressure of solution = 42.3 Torr + 108.81 Torr → 151.1 Torr