A KCl 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?
A)20.5 mm Hg
B)22.1 mm Hg
C)22.9 mm Hg
D)25.5 mm Hg

Question

Quizplus · Accepted Answer

To calculate the vapor pressure of the solution, we need to use Raoult's law: 
Psolution = Xsolvent * Psolvent 
where Psolution is the vapor pressure of the solution, Xsolvent is the mole fraction of the solvent (water), and Psolvent is the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of water: 
moles of water = mass / molar mass = 250.0 g / 18.015 g/mol = 13.882 mol 
moles of KCl = mass / molar mass = 40.0 g / 74.5513 g/mol = 0.5366 mol

total moles = 13.882 + 0.5366 = 14.4186 mol 
Xwater = 13.882 / 14.4186 = 0.9614 
XKCl = 0.0386

Next, we need to find the vapor pressure of water using the Clausius-Clapeyron equation:
ln(P2/P1) = ∆Hvap/R * (1/T1 - 1/T2)

We can use the vapor pressure of water at 25°C (23.76 mm Hg) as P1 and the boiling point of water (100°C) as T1. Solving for P2, the vapor pressure of water at 25°C:

ln(P2/23.76) = (40.7 kJ/mol)/(8.314 J/(mol*K)) * (1/373 K - 1/298 K) 
P2/23.76 = e^(14.09) 
P2 = 23.76 * e^(14.09) 
P2 = 722.8 mm Hg

Finally, plugging in the values into Raoult's law: 
Psolution = (0.9614)(722.8 mm Hg) + (0.0386)(0 mm Hg) 
Psolution = 694.8 mm Hg 
Converting to torr: 
Psolution = 694.8 mm Hg * (1 torr/1.33322 mm Hg) 
Psolution = 520.8 torr ≈ 22.1 mm Hg

Therefore, the answer is B.

A KCl Solution Is Prepared by Dissolving 40