Question

A solution contains 0.182 molmol NaClNaCl and 0.897 molH2OmolH2O. Calculate the vapor pressure of the solution at 55 ∘C∘C. The vapor pressure of pure water at 55 ∘C∘C is 118.1 torrtorr. (Assume that the solute completely dissociates.)

Answer 1

Answer:

Vapor pressure of solution is 78.2 Torr

Explanation:

This is solved by vapor pressure lowering:

ΔP =  P° . Xm . i

Vapor pressure of pure solvent  (P°) - vapor pressure of solution  = P° . Xm . i

NaCl  →  Na⁺  +  Cl⁻     i = 2

Let's determine the Xm (mole fraction) These are the moles of solute / total moles.

Total moles = moles of solvent + moles of solute

Total moles = 0.897 mol + 0.182 mol → 1.079 mol

0.182 / 1.079 = 0.168

Now we replace on the main formula:

118.1° Torr - P' = 118.1° Torr . 0.168 . 2

P' = - (118.1° Torr . 0.168 . 2 - 118.1 Torr)

P' =  78.2 Torr