The enthalpy of combustion of mol. Wt. 180 glucose is -2808 KJ `"mol"^(-1)` at `25^(@)`C . X and Y grams of glucose do you need to consume respectively cases [Assume wt=62.5 Kg].
(a) to climb a flight of stairs rising through 3M.
to climb a mountain of altitude 3000 M?
Assume that 25% of enthalpy can be converted to useful work.
X and Y are related as X=mY, then find m.

Explanation

To solve the problem, we need to calculate the amount of glucose required to climb a flight of stairs and a mountain, using the enthalpy of combustion of glucose. We will also find the relationship between the two amounts of glucose.

### Step-by-Step Solution:

1. **Given Data:**
- Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol
- Enthalpy of combustion of glucose = -2808 kJ/mol
- Weight of the person = 62.5 kg = 62500 g
- Gravitational acceleration (g) = 9.81 m/s²
- Useful work conversion efficiency = 25%

2. **Calculate the energy required to climb 3 meters:**
- The potential energy (PE) gained when climbing a height (h) is given by:
\[ PE = m \cdot g \cdot h \]
- For climbing 3 meters:
\[ PE = 62500 \, \text{g} \cdot 9.81 \, \text{m/s}^2 \cdot 3 \, \text{m} \]
\[ PE = 62500 \cdot 9.81 \cdot 3 = 1,843,125 \, \text{g m}^2/\text{s}^2 = 1,843.125 \, \text{kJ} \]

3. **Calculate the useful energy from glucose:**
- Since only 25% of the energy from glucose combustion is converted to useful work:
\[ \text{Useful energy} = 0.25 \times 2808 \, \text{kJ/mol} = 702 \, \text{kJ/mol} \]

4. **Calculate the moles of glucose needed (X) to climb 3 meters:**
- To find the moles of glucose required:
\[ \text{Moles of glucose} = \frac{\text{Total energy required}}{\text{Useful energy per mole}} \]
\[ \text{Moles of glucose} = \frac{1843.125 \, \text{kJ}}{702 \, \text{kJ/mol}} \approx 2.62 \, \text{mol} \]

5. **Convert moles to grams (X):**
- Using the molar mass of glucose:
\[ X = 2.62 \, \text{mol} \times 180 \, \text{g/mol} \approx 471.6 \, \text{g} \]

6. **Calculate the energy required to climb 3000 meters:**
- For climbing 3000 meters:
\[ PE = 62500 \, \text{g} \cdot 9.81 \, \text{m/s}^2 \cdot 3000 \, \text{m} \]
\[ PE = 62500 \cdot 9.81 \cdot 3000 = 1,836,375,000 \, \text{g m}^2/\text{s}^2 = 1,836,375 \, \text{kJ} \]

7. **Calculate the useful energy from glucose for 3000 meters:**
- Useful energy remains the same:
\[ \text{Useful energy} = 702 \, \text{kJ/mol} \]

8. **Calculate the moles of glucose needed (Y) to climb 3000 meters:**
- To find the moles of glucose required:
\[ \text{Moles of glucose} = \frac{1836375 \, \text{kJ}}{702 \, \text{kJ/mol}} \approx 2615.5 \, \text{mol} \]

9. **Convert moles to grams (Y):**
- Using the molar mass of glucose:
\[ Y = 2615.5 \, \text{mol} \times 180 \, \text{g/mol} \approx 471,000 \, \text{g} \]

10. **Find the relationship between X and Y:**
- Given that \( X = mY \):
\[ m = \frac{X}{Y} = \frac{471.6}{471000} \approx 0.001 \]

### Final Answer:
- The value of \( m \) is approximately **0.001**.