To 500 cm³ of water, 3.0 × 10⁻³ kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression at the freezing point? \(K_f\) and density of water are 1.86 K kg⁻¹ mol⁻¹ and 0.997 g cm⁻³ respectively. 

0.229

The correct answer is option 3. 0.229.

Mass of solute \(\text{ = 3.0 ×  }10^{−3} = 3.0 g\)

Molecular mass of the solute (Acetic acid) \(\text{= 60}\)

Mass of the solvent \(\text{ 500 ×  0.997 = 498.5 g}\)

We know that,

Degree of dissociation \((\alpha) = \frac{i − 1}{n − 1}\)

\(\text{or,                      }0.23 = \frac{i − 1}{2 − 1}\)

\(\text{or,                          }i  = 1.23\)

Now \(\Delta T = i × K_f × molality\)

⇒ \(\Delta T =1.23 × 1.86 × \frac{3 × 1000}{60 × 498.5}\)

⇒ \(\Delta T = \frac{1.23 × 93}{498.5}\)

∴ \(\Delta T = 0.229\)