In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is ________ $\times 10^{-2}$. (Round off to the Nearest Integer).
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The van't Hoff factor (i) quantifies the effect of a solute on colligative properties. For a non-electrolyte, i=1. For dissociation, i>1. For association, i<1. When both occur, calculate the final number of particles from each process separately and sum them up.
Let's assume we start with 1 mole of the acid HA.
According to the problem, 50% of the acid dimerizes.
Initial moles for dimerization = 0.5 mol.
The dimerization reaction is: 2HA $\rightleftharpoons$ (HA)$_2$.
The number of moles of dimer formed = $\frac{\text{Initial moles of HA}}{2} = \frac{0.5}{2} = 0.25$ mol.
The remaining 50% of the acid dissociates.
Initial moles for dissociation = 0.5 mol.
The dissociation reaction is: HA $\rightleftharpoons$ H$^+$ + A$^-$.
Assuming complete dissociation ("the rest dissociates"), 0.5 mol of HA will produce 0.5 mol of H$^+$ and 0.5 mol of A$^-$.
Total moles of particles from dissociation = 0.5 + 0.5 = 1.0 mol.
Now, calculate the total number of moles of all particles in the solution at equilibrium.
Total final moles = (moles of dimer) + (moles from dissociation)
Total final moles = 0.25 mol + 1.0 mol = 1.25 mol.
The van't Hoff factor (i) is defined as the ratio of the total moles of particles after association/dissociation to the initial moles of solute.
$i = \frac{\text{Total final moles}}{\text{Initial moles}}$.
$i = \frac{1.25}{1} = 1.25$.
The question asks for the answer in the format ________ $\times 10^{-2}$.
$1.25 = 125 \times 10^{-2}$.
The value to be filled in is 125.
