A carboxylic acid (RCOOH) undergoes 88% dimerization in benzene. What…

The correct answer is Option (1) → 0.56

The van’t Hoff factor i compares the number of particles actually present in solution with the number expected:

    \( \displaystyle i=\frac{\text{number of particles after association}}{\text{number of particles before association}} \)

Carboxylic acids associate into dimers in benzene through hydrogen bonding:

    \( \displaystyle \ce{2RCOOH <=> (RCOOH)2} \)

Start with 1 mol of acid and let a fraction α of it dimerise. Then (1 − α) mol stays as single molecules, and the associated part forms α/2 mol of dimers:

    \( \displaystyle i=\frac{(1-α)+\frac{α}{2}}{1}=1-\frac{α}{2} \)

Here the acid is 88% dimerised, so α = 0.88:

    \( \displaystyle i=1-\frac{0.88}{2} \)

    \( \displaystyle i=1-0.44 \)

    \( \displaystyle i=0.56 \)

Shortcut: Association always gives i < 1, and for a dimer i = 1 − α/2 = 1 − 0.44.