Chemistry, 24.06.2019 04:30 182075

Akcl solution is prepared by dissolving 40.0 g kcl in 250.0 g of water at 25°c. what is the vapor pressure of the solution if the vapor pressure of water at 25°c is 23.76 mm hg?

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Raoult's law states that for a given solution, the partial pressure of each component is equal to the mole fraction in that solution. The vapor pressure of the solution is 22 mmHg.

The vapor pressure is defined as the force exerted by the vapors in the walls of a container. It is a colligative property, such that the amount of substance increased or decreased is dependent on the amount of solute present.

The equation can be represented as:

P $_{\text{solvent}}$ - P $_{\text{solution}}$ = $\Delta$ P = X $_{\text{solute}}$ x P $_{\text{solvent}}$

where,

P $_{\text{solvent}}$ = vapor pressure of the pure solventP $_{\text{solution}}$ = vapor pressure of the solutionX $_{\text{solute}}$ = molar fraction of the solute

Now, calculating the mole fraction, where:

Moles of ions = 2 × number of moles of KCl = 1.134 molNumber of moles KCl = $\dfrac{40}{74.55}&= 0.567$ molesTotal moles are: 1.134 mol + 13.877 mol = 15.011 molX $_{\text{solute}}$ = $\dfrac{1.134}{15.01}&= 0.075$

Now, the vapor pressure of the solution can be calculated as:

$\Delta$ P = X $_{\text{solute}}$ x P $_{\text{solvent}}$

$\Delta$ P = 0.0755 × 23.76 mmHg = 1.78 mmHg

Hence, the vapor pressure of the solution:

P $_{\text{solution}}$ = $\Delta$ P + P $_{\text{solvent}}$

P $_{\text{solution}}$ = 23.76 mmHg - 1.79 mm Hg = 21.97mmHg

Therefore, the vapor pressure of the solution is approximately 22 mmHg.

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Answer from: Rch64

22.0 mmHg

Explanation:

The vapor pressure of a solution is a colligative property, which means that it is determined by the number of particles (molecules or ions) of solute present in a solution.

Raoult's law permits the calculations of the change of the vapor pressure of a solvent when a solute is added.

The equation is:

P solvent - P solution = ΔP = X solute × P solven

Where:

P solvent = vapor pressure of the pure solvent.P solution = vapor pressure of the solutionX solute = molar fraction of the solute

In the case of ionic solutes, you must take into account the number of ions that result from the ionization.

Calculating the molar fraction:

number of moles = mass in grams / molar massnumber of moles of KCl: 40.0 g / 74.5513 g/mol = 0.567 molmoles of ions = 2× number of moles of KCl = 1.134 molmoles of water: 250.0g / 18.015 g/mol = 13.877 moltotal moles = 1.134 mol + 13.877 mol = 15.011 molX solute = moles of ions / total moles = 1.134 mol / 15.011 mol = 0.0755

Calculating the change in the vapor pressure of the solution:

ΔP = X solute × P solvent = 0.0755 × 23.76 mmHg = 1.78 mmHg

Vapor pressure of the solution:

P solution = P solvent + ΔP = 23.76 mmHg - 1.79 mm Hg = 21.97mmHg

Rounding to three significant figures (because 40.0g has three significant figures): 22.0 mmHg ← answer.

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