Question

During anaerobic glycolysis in muscle, each mole of glucose is converted into two moles of lactate, yielding two moles of ATP. Assume molar mass of glucose is \(180\ \text{g/mol}\). If \(1\ \text{kg}\) of glucose undergoes complete conversion to lactate, the number of moles of ATP produced is _ _ _. (round off to one decimal place)

Hint:

In anaerobic glycolysis, each glucose molecule produces only \(2\) ATP molecules because pyruvate is converted to lactate instead of entering oxidative phosphorylation.

Answer 1

Correct Answer: 11.1

Step 1: Write the given information.
Mass of glucose:
\[ 1\ \text{kg}=1000\ \text{g} \] Molar mass of glucose:
\[ 180\ \text{g/mol} \] During anaerobic glycolysis:
\[ 1\ \text{mole glucose} \rightarrow 2\ \text{moles ATP} \]

Step 2: Calculate the number of moles of glucose.
\[ \text{Moles of glucose} = \frac{\text{Mass}}{\text{Molar mass}} \] \[ = \frac{1000}{180} \] \[ =5.555\ldots\ \text{mol} \]

Step 3: Determine ATP production.
Each mole of glucose produces \(2\) moles of ATP.
Therefore, total ATP produced is
\[ 5.555\ldots \times 2 \] \[ =11.111\ldots \]

Step 4: Round off to one decimal place.
\[ 11.111\ldots \approx 11.1 \]

Step 5: Final conclusion.
Hence, the number of moles of ATP produced is
\[ \boxed{11.1} \]