QUESTION 1 1 points A KCI solution is prepared by dissolving 60.0 g KCl in 250.0

Question

QUESTION 1 1 points A KCI solution is prepared by dissolving 60.0 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? 21.3 mm Hg 22.5 mm Hg 20.5 mm Hg 25.5 mm Hg QUESTION 2 1 points An aqueous solution has a normal boiling point of 105.0°C. What is the freezing point of this solution? For water Kb is 0.51°C/m and Kf 1.86°C/m -14.7°C 18.2°C 14.7°C 18.2°C 0 QUESTION 3 1 points The following diagram shows a close-up view of the vapor pressure curves for a pure solvent and a solution containing a nonvolatile solute dissolved in this solvent (a) (b Pressure (mm Hg) Temperature (°c) Which curve is the solvent and what happens to the vapor pressure when the solute is dissolved in the solvent? Curve (a) is the solvent and the vapor pressure decreases Curve (b) is the solvent and the vapor pressure increases Curve (a) is the solvent and the vapor pressure increases Curve (b) is the solvent and the vapor pressure decreases

Explanation / Answer

Ans 1

Mass of KCl = 60g

Moles of KCl = mass/molecular weight

= 60 g / 74.5513 g/mol

= 0.8048 mol

Mass of water = 250 g

Moles of water = 250g / 18g/mol

= 13.888 mol

Total Moles = 0.8048 + 13.888 = 14.6936 mol

Mol fraction of water = moles of water / total moles

= 13.888/14.6936

= 0.9452

Vapor pressure of water at 25°C = 23.76 mmHg

From the Raoult’s law

Vapor pressure of solution = mol fraction of water x Vapor pressure of water

= 0.9452 x 23.76

= 22.5 mmHg

Option B is the correct answer

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.