Please answer the question with proper solution.

We have, w = 2 g, W = 25 g, ΔT = 1.62, K’_f = 4.9

Using ΔT = (1000 × K’_f × w)/(m × W)

1.62 = (1000 × 4.9 × 2)/(25 × m)

Therefore, m_exp = 241.98

Now, for nC₆H₅COOH ↔ (C₆H₅COOH)_n

Before association 0

1 – α/n

Therefore, total number of mole at equilibrium = 1 – α + α/n

Or, m_N/m_exp = 1 – α + α/n

For dimer formation, n = 2

Or, 122.0/241.98 = 1 – α + α/2

[For C₆H₅COOH, m_N = 122.0]

Or, 1 – α/2 = 0.504

Or, α = 0.992

= 99.2%