Question

A 0.1 M solution of NaCl is found to be isotonic with a 1% urea solution. Calculate:

1. van’t Hoff factor, and
2. degree of dissociation of NaCl.

Answer 1

Given: A 0.1 M NaCl solution is isotonic with a 1% urea solution

Molar mass of urea = 60 g/mol

Volume of solution assumed = 100 mL → 1 g urea in 100 mL = 1%

Therefore, moles of urea = \[\ce{\frac{1}{60}}\] mol in 0.1 L →

\[\ce{C_{urea} = \frac{1}{60 \times 0.1} = 0.1667}\] mol/L

i. Using osmotic pressure equation

Since both solutions are isotonic at the same temperature:

π_NaCl = π_urea 

⇒ i_NaCl × C_NaCl

= 1 × C_urea

Substitute:

i × 0.1 = 0.1667

⇒ \[\ce{i = \frac{0.1667}{0.1}}\]

= 1.67

ii. Degree of dissociation (α):

NaCl dissociates as 

\[\ce{NaCl -> Na+ + Cl-}\]

⇒ n = 2

By using the formula, 

i = 1 + α(n − 1) = 1 + α(2 − 1) = 1 + α

1 + α = 1.67     ...(i = 1.67)

⇒ α = 1.67 − 1

α = 0.67

∴ The van’t Hoff factor (i) is 1.67, and the degree of dissociation is 0.67.