Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
The electron equivalent per liter (e -eq L-1) of acetate ion solution of concentration
10 g L-1 is _______ e-eq L-1. (rounded off to one decimal place)
Official Answer Key:
1.3 to 1.4
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2026 Q.57)**
**1. Governing Concept & Definition**:
- **Topic**: Environmental Biotechnology & Electron Equivalents
- For acetate ($\text{CH}_3\text{COO}^-$), the oxidation half-reaction is:
$\text{CH}_3\text{COO}^- + 2\text{H}_2\text{O} \rightarrow 2\text{CO}_2 + 7\text{H}^+ + 8e^-$
Thus, 1 mole of acetate ion corresponds to $8\text{ electron equivalents}$ ($8\text{ e-eq}$).
**2. Detailed Step-by-Step Calculation**:
- **Molecular Weight of Acetate ($\text{CH}_3\text{COO}^-$)**:
$\text{MW} = 12(2) + 1(3) + 16(2) = 24 + 3 + 32 = 59\text{ g mol}^{-1}$
- **Molar Concentration of $10\text{ g L}^{-1}$ Acetate**:
$\text{Molarity} = \frac{10\text{ g L}^{-1}}{59\text{ g mol}^{-1}} \approx 0.16949\text{ mol L}^{-1}$
- **Electron Equivalent per Liter**:
$\text{e-eq L}^{-1} = 0.16949\text{ mol L}^{-1} \times 8\text{ e-eq mol}^{-1} \approx 1.356\text{ e-eq L}^{-1} \approx 1.4\text{ e-eq L}^{-1}$
**3. Final Answer**:
- **Calculated Value**: **1.4 e-eq L⁻¹**
- **Official GATE Accepted Range**: **1.3 to 1.4**
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