1. Tardigrade
  2. Question
  3. Chemistry
  4. Determine the degree of association (polymerization) for the reaction in aqueous solution 6 textHCHO leftharpoons textC6 textH1 2 textO6 If observed (mean) molar mass of textHCHO and textC text6 textH text12 textO text6 is 150:

Q. Determine the degree of association (polymerization) for the reaction in aqueous solution
$6 \text{HCHO} \rightleftharpoons \text{C}_{6} \text{H}_{1 2} \text{O}_{6}$
If observed (mean) molar mass of $\text{HCHO}$ and $\text{C}_{\text{6}} \text{H}_{\text{12}} \text{O}_{\text{6}}$ is 150:

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Solution:

$ & 6\text{HCHO} & \rightleftharpoons & \left(\text{C}\right)_{6}\left(\text{H}\right)_{1 2}\left(\text{O}\right)_{6} \\ \text{Conc. at equilibrium} & \text{C}\left(1 - \text{α}\right) & & \frac{\text{C} \text{α}}{6}$
$\text{α}=\frac{\left(\text{M}\right)_{\text{T} \text{h}} - \left(\text{M}\right)_{\text{o} \text{b} \text{s}}}{\left(\right. \text{n} - 1 \left.\right) \left(\text{M}\right)_{\text{o} \text{b} \text{s}}} \\ \\ = \, \, \frac{30 - 150}{\left[\right. \frac{1}{6} - 1 \left]\right. 150} \\ \\ = \, \, \frac{120}{\frac{5}{6} \times 150}=0.96 \\ $

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