Given that the vapor pressure of water is 17.54 torr at 20 °c, calculate the vapor-pressure lowering of aqueous solutions that are 2.10 m in (a) sucrose, c12h22o11, and (b) sodium chloride. assume 100% dissociation for electrolytes.

0.640 Torr; 1.28 Torr  

Step-by-step explanation:

According to Raoult's Law,

(1) p₁ = χ₁p₁°,  

where p₁ is the solvent

χ₁ + χ₂ = 1                        Subtract χ₁ from each side

      χ₂ = 1 – k₁                          Substitute in (1)

(2)    p₁ = (1 – χ₂)p₁°    Remove parentheses

       p₁ = p₁°- χ₂p₁°      Subtract p₁° from each side

p₁ - p₁° = -χ₂p₁°              Multiply each side by -1

     Δp =   χ₂p₁°

Δp is the vapour pressure lowering.

If the solute is an electrolyte, we must insert the van't Hoff i factor.

ΔP = iχ₂P₁°

(1) Calculate χ₂ for each solution

Water:

1000 g = 1000 × 1/18.02

          = 55.49 mol

      χ₂ = 2.10/(2.10 + 55.49)

          = 2.10/57.59

          = 0.036 46

(2) Calculate the vapour pressure lowering

(a) Sucrose

   i = 1

ΔP = 1 × 0.036 46 × 17.54

ΔP = 0.640 Torr

(b) NaCl

   i = 2

ΔP = 2 × 0.640

ΔP = 1.28 torr

it is a good conductor

explanation:

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