The Van't Hoff factor of BaCl_2 at a concentr | vedclass

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(A) The dissociation reaction for $BaCl_2$ is: $BaCl_2 \rightleftharpoons Ba^{2+} + 2Cl^-$Let the initial moles be $1$. After dissociation,the moles a

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$49$

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(A) The dissociation reaction for $BaCl_2$ is: $BaCl_2 ightleftharpoons Ba^{2+} + 2Cl^-$Let the initial moles be $1$. After dissociation,the moles are: $BaCl_2 = 1 - \alpha$,$Ba^{2+} = \alpha$,$Cl^- = 2\alpha$,where $\alpha$ is the degree of dissociation.The total number of particles after dissociation is $n = (1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha$.The Van't Hoff factor $i$ is given by the ratio of total particles to initial particles: $i = 1 + 2\alpha$.Given $i = 1.98$,we have: $1.98 = 1 + 2\alpha$.$2\alpha = 1.98 - 1 = 0.98$.$\alpha = 0.49$.Percentage dissociation = $\alpha 	imes 100 = 0.49 	imes 100 = 49\%$.

The Van't Hoff factor of $BaCl_2$ at a concentration of $0.01 \ M$ is $1.98$. The percentage dissociation of $BaCl_2$ at this concentration will be: (in $\%$)

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