If BaCl_2 ionizes to an extent of 80 % in aqu | vedclass

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(A) The dissociation reaction for $BaCl_2$ is: $BaCl_{2(aq)} \rightarrow Ba^{2+}_{(aq)} + 2Cl^-_{(aq)}$Initially, we have $1$ mole of $BaCl_2$.Given t

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$2.6$

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(A) The dissociation reaction for $BaCl_2$ is: $BaCl_{2(aq)} ightarrow Ba^{2+}_{(aq)} + 2Cl^-_{(aq)}$Initially, we have $1$ mole of $BaCl_2$.Given the degree of dissociation $\alpha = 80\% = 0.8$.At equilibrium, the moles are:$BaCl_2 = 1 - \alpha = 1 - 0.8 = 0.2$$Ba^{2+} = \alpha = 0.8$$Cl^- = 2\alpha = 2 	imes 0.8 = 1.6$Total moles at equilibrium = $0.2 + 0.8 + 1.6 = 2.6$The van't Hoff factor $i$ is defined as the ratio of total moles after dissociation to the initial moles:$i = \frac{2.6}{1} = 2.6$

If $BaCl_2$ ionizes to an extent of $80 \%$ in aqueous solution, the value of van't Hoff factor is

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