JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Solutions

In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is

____

\times 1 0^{- 2}

. (Round off to the Nearest Integer).

Answer: 125

Solution

Take

1

mole of HA. Half of it,

0.5

mole, dimerises as

2 HA \to (HA)_{2}

and yields

0.25

mole of dimer. The remaining

0.5

mole dissociates as

HA \to H^{+} + A^{-}

and yields

0.5

mole of each ion. No undissociated monomer is left, so the particles present total

0.25 + 0.5 + 0.5 = 1.25 mole

The van't Hoff factor is the ratio of observed particles to the particles taken,

i = \frac{1.25}{1} = 1.25

so the van't Hoff factor is

125 \times 1 0^{- 2}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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