Formaldehyde polymerizes to form glucose according to the reaction, `6HCHO rarr C_(6)H_(12)O_(6)` The theoretically computed equilibrium constant for this reaction is found to be `6 xx 10^(22)` If 1M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be :

Options

A. `1.6 xx 10^(-2)M`
B. `1.6 xx 10^(-4)M`
C. `1.6 xx 10^(-6)M`
D. `1.6 xx 10^(-8)M`

Explanation

To solve the problem, we need to find the concentration of formaldehyde (HCHO) in a 1M solution of glucose (C₆H₁₂O₆) that dissociates according to the given equilibrium reaction:

\[ 6 \text{HCHO} \rightleftharpoons \text{C}_{6}\text{H}_{12}\text{O}_{6} \]

The equilibrium constant \( K \) for this reaction is given as \( 6 \times 10^{22} \).

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction shows that 6 moles of formaldehyde polymerize to form 1 mole of glucose. This means that for every mole of glucose formed, 6 moles of formaldehyde are consumed.

2. **Set Up the Equilibrium Expression**:
The equilibrium constant \( K \) for the reaction can be expressed as:
\[ K = \frac{[\text{C}_{6}\text{H}_{12}\text{O}_{6}]}{[\text{HCHO}]^6} \]
Here, \([\text{C}_{6}\text{H}_{12}\text{O}_{6}]\) is the concentration of glucose, and \([\text{HCHO}]\) is the concentration of formaldehyde.

3. **Substitute Known Values**:
We know that the concentration of glucose is 1M:
\[ K = \frac{1}{[\text{HCHO}]^6} \]
Given \( K = 6 \times 10^{22} \), we can substitute this into the equation:
\[ 6 \times 10^{22} = \frac{1}{[\text{HCHO}]^6} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ [\text{HCHO}]^6 = \frac{1}{6 \times 10^{22}} \]

5. **Calculating \([\text{HCHO}]\)**:
Taking the sixth root of both sides:
\[ [\text{HCHO}] = \left(\frac{1}{6 \times 10^{22}}\right)^{\frac{1}{6}} \]
This simplifies to:
\[ [\text{HCHO}] = \frac{1}{(6)^{\frac{1}{6}} \times (10^{22})^{\frac{1}{6}}} \]

6. **Calculating the Values**:
First, calculate \( (10^{22})^{\frac{1}{6}} = 10^{\frac{22}{6}} = 10^{3.6667} \approx 4641.59 \).
Now calculate \( (6)^{\frac{1}{6}} \approx 1.348 \).

Therefore:
\[ [\text{HCHO}] \approx \frac{1}{1.348 \times 4641.59} \approx \frac{1}{6256.86} \approx 0.000159 \]

7. **Final Result**:
Converting this to scientific notation:
\[ [\text{HCHO}] \approx 1.59 \times 10^{-4} \text{ M} \]

### Conclusion:
The concentration of formaldehyde in the solution is approximately \( 1.59 \times 10^{-4} \text{ M} \).