Some KCl is dissolved in water at 25^$\circ$C, where it completely dissociates. The vapor pressure of pure water at 25^$\circ$C is 28.3 mmHg. Estimate the mass in grams of KCl needed per liter of pure water to reduce the vapor pressure of water at 25^$\circ$C by 5%. A) 110 g B) 120 g C) 130 g D) 140 g E) None of the above

Question

Quizplus · Accepted Answer

To reduce the vapor pressure by 5%, the mole fraction of KCl must be such that the vapor pressure of the solution is 95% of the pure water's vapor pressure. Using Raoult's Law and the dissociation of KCl into K+ and Cl-, the calculation shows that approximately 110 g of KCl is needed per liter of water.

Some KCl Is Dissolved in Water at 25 ∘\circ∘ C, Where It Completely Dissociates

Some KCl Is Dissolved in Water at 25^$\circ$C, Where It Completely Dissociates