An athlete is given 100 g of glucose (`C_6H_(12)O_6`) for energy. This is equivalent to 1800 KJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event . In order to avoid storage of energy, the weight of extra water he would need to perspire is _______g (Nearest integer)Assume that there is no other way of consuming stored energyGiven : The enthalpy of evaporation of water is 45 KJ `mol^(–1)`Molar mass of C, H & O are 12, 1 and 16 g `mol^(–1)`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Determine the energy utilized by the athlete
The athlete is given 100 g of glucose, which provides 1800 kJ of energy. Since the athlete utilizes 50% of this energy for sports activities, we calculate:

\[ \text{Energy utilized} = 0.5 \times 1800 \, \text{kJ} = 900 \, \text{kJ} \]

### Step 2: Calculate the moles of water evaporated
The energy required for perspiration (evaporation of water) can be calculated using the enthalpy of evaporation of water, which is given as 45 kJ/mol. To find the number of moles of water evaporated, we use the formula:

\[ \text{Moles of water} = \frac{\text{Energy utilized}}{\text{Enthalpy of evaporation}} = \frac{900 \, \text{kJ}}{45 \, \text{kJ/mol}} = 20 \, \text{mol} \]

### Step 3: Calculate the weight of water evaporated
Next, we need to convert the moles of water into grams. The molar mass of water (H₂O) is approximately 18 g/mol. Therefore, the weight of water evaporated can be calculated as follows:

\[ \text{Weight of water} = \text{Moles of water} \times \text{Molar mass of water} = 20 \, \text{mol} \times 18 \, \text{g/mol} = 360 \, \text{g} \]

### Final Answer
The weight of extra water the athlete would need to perspire to avoid storage of energy is **360 g** (nearest integer).

---