Question

Determine the degree of association (polymerisation) for the reaction in aqueous solution $6HCHO \rightleftharpoons {C_6}{H_{12}}{O_6}$
If observed (mean) molar mass of ${\text{HCHO}}$ and ${C_6}{H_{12}}{O_6}$ is ${\text{150}}$.
A) ${\text{0}} \cdot {\text{50}}$
B) ${\text{0}} \cdot {\text{833}}$
C) ${\text{0}} \cdot {\text{90}}$
D) ${\text{0}} \cdot {\text{96}}$

Answer 1

Hint: The degree of association can be elaborated as the fraction of the total number of molecules which gets combined together that is associated with the formation of a bigger molecule. One can get this ratio by using the formula of the mass value of components and total molar mass.

Complete step by step answer:
1) First of all we will learn what exactly has been asked and the concept of degree of association. The degree of association in simple words is a fraction value of how many molecules got combined that is associated to form a big compound.
2) So, in the given question it has been asked that to form a compound ${C_6}{H_{12}}{O_6}$ we need how much fraction of ${\text{HCHO}}$ has been used. Now let us write the reaction equation as below,
$6HCHO \rightleftharpoons {C_6}{H_{12}}{O_6}$
Now let assume a concentration of the ${\text{HCHO}}$ initial is $C(1 - \alpha )$ where $\alpha $ is the value of the degree of association and for ${C_6}{H_{12}}{O_6}$ it will be $\dfrac{{{C_\alpha }}}{6}$.
3) Now as we know the formula,
$\dfrac{{{\text{Observed molar concentration}}}}{{{\text{Initial molar concentration}}}} = \dfrac{{{M_T}}}{{{M_O}}}$
In the above equation, the value of ${M_T}$ is the molar mass of ${\text{HCHO}}$ which is as follows,
Molar mass of ${\text{HCHO}}$ $ = 1 + 12 + 1 + 16 = 30$
The value ${M_O}$ is the observed (mean) molar mass of ${\text{HCHO}}$ and ${C_6}{H_{12}}{O_6}$ is ${\text{150}}$.
4) Now lets put these values in the above formula we get,
$\dfrac{{C(1 - \alpha ) + \dfrac{{{C_\alpha }}}{6}}}{C} = \dfrac{{30}}{{150}}$
Now we can take the value of C common on the left hand side of the equation,
$\dfrac{{C\left[ {\left( {1 - \alpha } \right) + \dfrac{1}{6}} \right]}}{C} = \dfrac{{30}}{{150}}$
Now we can cancel out the value of C which is common and we get,
$(1 - \alpha ) + \dfrac{\alpha }{6} = \dfrac{{30}}{{150}}$
Now we can take $\alpha $ value common in the L.H.S. of the equation,
$\alpha \left( {\dfrac{1}{6} - 1} \right) = \dfrac{{30}}{{150}} - 1$
Now by calculating the above we get,
$\alpha \left( {\dfrac{5}{6}} \right) = \dfrac{3}{5} - 1$
$\alpha \left( {\dfrac{5}{6}} \right) = \dfrac{4}{5}$
Now taking the $\alpha $ value on one side we get,
$\alpha = \dfrac{4}{5} \times \dfrac{6}{5}$
$\alpha = \dfrac{{24}}{{25}} = 0 \cdot 96$
Therefore, the degree of association (polymerisation) for the reaction in aqueous solution is ${\text{0}} \cdot {\text{96}}$ which shows option D as the correct choice.

Note:
The degree of association is also determined by using the Van’t Hoff factor. The Van’t Hoff factor is the quantity of effect which is produced on the colligative properties and for associative liquids, it is less than one and for dissociation, its value is more than one.