A sol of AgI is prepared by mixing equal volu | vedclass

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(D) The reaction between $AgNO_{3}$ and $KI$ is: $AgNO_{3} + KI \longrightarrow AgI(s) + KNO_{3}$.Since equal volumes are mixed,the number of moles of

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Sol obtained is a negative sol with $I^{-}$ adsorbed on $AgI$.

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(D) The reaction between $AgNO_{3}$ and $KI$ is: $AgNO_{3} + KI \longrightarrow AgI(s) + KNO_{3}$.Since equal volumes are mixed,the number of moles of $KI$ $(0.2 \ M)$ is greater than the number of moles of $AgNO_{3}$ $(0.1 \ M)$.Thus,$KI$ is in excess.The $AgI$ particles preferentially adsorb the common ion present in excess from the dispersion medium.In this case,$I^{-}$ ions are in excess and are adsorbed on the surface of $AgI$ particles,resulting in a negatively charged sol.

$A$ sol of $AgI$ is prepared by mixing equal volumes of $0.1 \ M \ AgNO_{3}$ and $0.2 \ M \ KI$. Which of the following statements is correct?

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