The van't Hoff factor i for a compound which | vedclass

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(C) For a solute undergoing dissociation,the number of particles increases,which leads to an increase in the observed colligative property. Since $i =

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Greater than one and less than one

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(C) For a solute undergoing dissociation,the number of particles increases,which leads to an increase in the observed colligative property. Since $i = \frac{	ext{observed colligative property}}{	ext{normal colligative property}}$,the value of $i$ is greater than $1$ $(i > 1)$.In the case of association,the number of particles decreases,leading to a decrease in the observed colligative property. Therefore,the value of $i$ is less than $1$ $(i < 1)$.

The van't Hoff factor $i$ for a compound which undergoes dissociation in one solvent and association in other solvent is respectively

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What is association? What is dissociation?

If a $0.01 \ m$ aqueous solution of an electrolyte freezes at $-0.056 \ ^{\circ}C$,calculate the van't Hoff factor $(i)$ for the electrolyte. (Cryoscopic constant of water $K_{f} = 1.86 \ K \ kg \ mol^{-1}$)

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