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The number of moles of lead nitrate needed to coagulate a mol of colloidal $\;[AgI]I^{-}\;$ is

$(a)\;2\qquad(b)\;1\qquad(c)\;\large\frac{1}{2}\qquad(d)\;\large\frac{2}{3}$

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Answer : 1
Explanation :
$2 [AgI]I^{-} + Pb^{2+} \to PbI_{2} + 2AgI$
Thus 2 moles of $[AgI]I^{-}\;$ are coagulated by 1 mole of $\;Pb^{2+}\;$ i.e ; 1 mole of $\;Pb(NO_{3})_{2}$
answered Mar 18, 2014 by yamini.v
 

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