Determine the degree of association (polymerization) for the reaction in aqueous solution, if observed (mean) molar mass of HCHO and `C_(6)H_(12)O_(6)` is `150` :
`6HCHOhArrC_(6)H_(12)O_(6)`

Options

A. (a) `0.50`
B. (b) `0.833`
C. (c) `0.90`
D. (d) `0.96`

Answer

To determine the degree of association (polymerization) for the reaction \(6 \text{HCHO} \rightleftharpoons \text{C}_6\text{H}_{12}\text{O}_6\) in aqueous solution, we can follow these steps:

### Step 1: Write the reaction and identify the components
The reaction is given as:
\[ 6 \text{HCHO} \rightleftharpoons \text{C}_6\text{H}_{12}\text{O}_6 \]
We have 6 moles of formaldehyde (HCHO) associating to form one mole of glucose (\(C_6H_{12}O_6\)).

### Step 2: Define the initial concentration
Let the initial concentration of HCHO be \(C\). At equilibrium, some of the HCHO will have reacted to form glucose.

### Step 3: Define the degree of association
Let \(\alpha\) be the degree of association. The concentration of HCHO that remains unreacted at equilibrium will be:
\[ C(1 - \alpha) \]
The concentration of glucose formed will be:
\[ \frac{C\alpha}{6} \]

### Step 4: Write the expression for the total concentration at equilibrium
At equilibrium, the total concentration of the solution will be the sum of the concentrations of unreacted HCHO and the formed glucose:
\[ \text{Total concentration} = C(1 - \alpha) + \frac{C\alpha}{6} \]

### Step 5: Simplify the total concentration
Factoring out \(C\), we get:
\[ \text{Total concentration} = C \left(1 - \alpha + \frac{\alpha}{6}\right) \]
This can be simplified to:
\[ \text{Total concentration} = C \left(1 - \alpha + \frac{\alpha}{6}\right) = C \left(1 - \frac{5\alpha}{6}\right) \]

### Step 6: Calculate the theoretical molar mass
The molar mass of HCHO is calculated as follows:
\[ \text{Molar mass of HCHO} = 1 + 12 + 1 + 16 = 30 \text{ g/mol} \]
The theoretical molar mass when 6 moles of HCHO react to form 1 mole of \(C_6H_{12}O_6\) is:
\[ \text{Theoretical molar mass} = \frac{6 \times 30}{1} = 180 \text{ g/mol} \]

### Step 7: Use the observed molar mass to find the degree of association
We are given that the observed molar mass \(M_O\) is 150 g/mol. The degree of polymerization (association) can be calculated using the formula:
\[ \text{Degree of association} = \frac{M_T}{M_O} \]
Where \(M_T\) is the theoretical molar mass. Thus:
\[ \text{Degree of association} = \frac{180}{150} = \frac{6}{5} = 1.2 \]

### Step 8: Relate the degree of association to \(\alpha\)
The degree of association can also be expressed in terms of \(\alpha\):
\[ \text{Degree of association} = 1 - \frac{5\alpha}{6} \]
Setting this equal to \(\frac{6}{5}\):
\[ 1 - \frac{5\alpha}{6} = \frac{1}{5} \]

### Step 9: Solve for \(\alpha\)
Rearranging gives:
\[ \frac{5\alpha}{6} = 1 - \frac{1}{5} \]
\[ \frac{5\alpha}{6} = \frac{4}{5} \]
Multiplying both sides by 6:
\[ 5\alpha = \frac{24}{5} \]
\[ \alpha = \frac{24}{25} = 0.96 \]

### Conclusion
Thus, the degree of association (polymerization) is:
\[ \alpha = 0.96 \]