Question

A 0.15 M aqueous solution of KCl exerts an osmotic pressure of 6.8 atm at 310 K. Calculate the degree of dissociation of KCl. (R = 0.0821 L atm K⁻¹ mol⁻¹). 

Answer 1

R = 0.0821 L atm K⁻¹ mol⁻¹ 

π = i = 1.78

6.8 = i × 0.15 × 0.0821 × 310

i = 1.78

`alpha = ("i" - 1)/("n" - 1)`   ...[For KCl, n = 2]

`alpha = (1.78 - 1)/(2 - 1)`

Degree of dissociation, α = 0.78 or 78 %

Answer 2

Given: Molarity (C) = 0.15 mol/L

Osmotic pressure (π) = 6.8 atm

Temperature (T) = 310 K

Gas constant R = 0.0821 L atm mol⁻¹ K⁻¹

Use van’t Hoff equation

π = iCRT

`i = pi/(CRT)`

= `6.8/(0.15 xx 0.0821 xx 310)`

= `6.8/3.81765`

= 1.78

\[\ce{KCl -> K+ + Cl-}\]

i = 1 + α

1.78 = 1 + α

α = 1.78 − 1

α = 0.78

α = 0.78 or 78%