Determine the degree of association (polymeri | vedclass

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(D) The reaction is $6 HCHO \rightleftharpoons C_6H_{12}O_6$. Let the initial concentration of $HCHO$ be $c$. At equilibrium,the concentration of $HCH

Vedclass · Accepted Answer

$0.96$

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(D) The reaction is $6 HCHO ightleftharpoons C_6H_{12}O_6$. Let the initial concentration of $HCHO$ be $c$. At equilibrium,the concentration of $HCHO$ is $c(1-\alpha)$ and the concentration of $C_6H_{12}O_6$ is $\frac{c\alpha}{6}$. The total number of moles at equilibrium is $n_{total} = c(1-\alpha) + \frac{c\alpha}{6} = c(1 - \frac{5\alpha}{6})$. The van't Hoff factor $i$ is given by $\frac{n_{total}}{n_{initial}} = \frac{c(1 - \frac{5\alpha}{6})}{c} = 1 - \frac{5\alpha}{6}$. Also,$i = \frac{M_{theoretical}}{M_{observed}} = \frac{30}{150} = 0.2$. Equating the two expressions: $1 - \frac{5\alpha}{6} = 0.2$. $\frac{5\alpha}{6} = 0.8$. $\alpha = \frac{0.8 	imes 6}{5} = 0.96$.

Determine the degree of association (polymerization) for the reaction in aqueous solution $6 HCHO \rightleftharpoons C_6H_{12}O_6$. If the observed (mean) molar mass of the mixture is $150 \ g/mol$. (Given: Molar mass of $HCHO = 30 \ g/mol$)

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