An NaCl solution is prepared by dissolving 20.0 g NaCl in 150.0 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.56 torr? Group of answer choices

Vapor pressure for the solution is 21.70 Torr

Explanation:

Vapor pressure lowering → Colligative property

P° = Vapor pressure of pure solvent

P' = Vapor pressure of solution

Xm = Mole fraction of solute → Moles of solute / Total moles

Total moles = Moles of solute + moles of solvent

Formula = P° - P' = P° . Xm . i

Let's determine the moles of solute and solvent

20 g  / 58.45 g/mol = 0.342 moles NaCl

150 g / 18 g/mol = 8.333 moles H₂O

Total moles → 8.333 + 0.342 = 8.675 moles

Mole fraction of solute = 0.342 / 8.675 = 0.039

i → NaCl ⇒ Na⁺ + Cl⁻   i = 2 (Van't Hoff factor, number of ions dissolved)

Let's replace the data, that was obtained from the problem.

23.56 Torr - P' = 23.56 Torr . 0.039 . 2

P' = - (23.56 Torr . 0.039 . 2 - 23.56 Torr) → 21.70 Torr

the answer is 1 i believe