Question

Use the data in the following table for three different aqueous solutions of CaCl₂ to calculate the apparent value of the Van’t Hoff factor.

Molality

Freezing-Point

Depression

(8˚C)

0.0225 0.110

0.0910 0.440

0.278 1.330

Short answer

CaCl₂ ----- Ca⁺² + 2Cl ^–

CaCl₂ having 0.0225 molality i = 1.045

CaCl₂ having 0.0910 molality i = 1.182

CaCl₂ having 0.278 molality i = 1.556

Step-by-step solution

Step 1:

Van’t Hoff factor is expressed as i = 1+ (y-1) x

y = number of dissociated ions

x = % value

i = 1+ (y-1) x for 0.0225 molality

i = 1 + (3-1) 0.0225

i = 1.045

i = 1+ (y-1) x for 0.0910 molality

i = 1 + (3-1) 0.0910

i = 1.182

i = 1+ (y-1) x for 0.278 molality

i = 1 + (3-1) 0.278

i = 1.556