The vapor pressure of water at 45.0 ^$\circ$C is 71.93 mmHg. Calculate the vapor pressure of a solution of 1.50m K₂CO₃ at this temperature (assume 100% dissociation). A) 1.89 mmHg B) 5.39 mmHg C) 66.5 mmHg D) 70.0 mmHg E) 108 mmHg

Question

Quizplus · Accepted Answer

First, calculate the vapor pressure of pure water at 45.0°F using the given value:
P°water = 71.93 mmHg

Then, use Raoult's Law to calculate the vapor pressure of the solution:
Pwater = Xwater * P°water
PKSbO = XKSbO * PKSbO°

Since the compound is assumed to be 100% dissociated, the mole fraction of KSbO in the solution is equal to its concentration:
XKSbO = 1.50 mol/L / (1.50 mol/L + 55.5 mol/L) = 0.0266

The vapor pressure of the solution is then:
Ptotal = Pwater * Xwater + PKSbO * XKSbO
Ptotal = (1 - XKSbO) * Pwater + XKSbO * PKSbO° = (1 - 0.0266) * 71.93 mmHg + 0.0266 * 76.41 mmHg = 66.5 mmHg

Therefore, the correct answer is C.

The Vapor Pressure of Water at 45 ∘\circ∘ C Is 71

The Vapor Pressure of Water at 45 $\circ$ C Is 71