For a 0.1 M Ba(NO_3)_2 solution,the van't Ho | vedclass

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(B) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-$. Here,the number of ions produced per formula unit,$n = 3

Vedclass · Accepted Answer

$87$

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(B) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 ightarrow Ba^{2+} + 2NO_3^-$. Here,the number of ions produced per formula unit,$n = 3$. The formula for the van't Hoff factor $(i)$ is: $i = 1 + (n - 1)\alpha$,where $\alpha$ is the degree of dissociation. Substituting the given values: $2.74 = 1 + (3 - 1)\alpha$. $2.74 = 1 + 2\alpha$. $2\alpha = 1.74$. $\alpha = 0.87$. Therefore,the degree of dissociation in percentage is $0.87 	imes 100 = 87\%$.

For a $0.1 \ M$ $Ba(NO_3)_2$ solution,the van't Hoff factor is $2.74$. The degree of dissociation is ....... $\%$.

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