Relative Lowering of Vapor Pressure Concept

Relative lowering of vapor pressure is a colligative property that depends on the number of solute particles in solution. For electrolytes that dissociate, we must account for the total number of ions produced.

Step 1: Determine the van't Hoff factor (i)

KBr dissociates as: $\mathrm{KBr}\to K^{+}+{\mathrm{Br}}^{-}$

Given 20% dissociation: α = 0.2

van't Hoff factor: $i=1+\alpha (n-1)$ where n = 2

$i=1+0.2(2-1)=1.2$

Step 2: Calculate effective molality

Effective molality = i × given molality = 1.2 × 0.1 = 0.12 m

Step 3: Calculate relative lowering of vapor pressure

For dilute aqueous solutions: $\frac{\mathrm{\Delta P}}{P^0}=\frac{i\times m}{55.56}$

Where 55.56 is the molality of pure water (1000g/18g/mol)

$\frac{\mathrm{\Delta P}}{P^0}=\frac{0.12}{55.56}$

Final Answer

The relative lowering of vapor pressure is $\frac{0.12}{55.56}$

Related Topics

• Colligative Properties
• Van't Hoff Factor
• Electrolyte Solutions
• Degree of Dissociation

Important Formulae

Relative lowering of vapor pressure: $\frac{\mathrm{\Delta P}}{P^0}=\frac{n}{N+n}\approx \frac{n}{N}$

Van't Hoff factor: $i=1+\alpha (n-1)$

For dilute solutions: $\frac{\mathrm{\Delta P}}{P^0}=\frac{i\times m}{55.56}$