A boy after swimming comes out from a pool covered with a film of water weighing 80 g .How much heat must be supplied to evaporate this water ? `(Delta_(v) H^(@)=40.79 kJmol^(-1))`

Options

A. `1.61 xx10^(2) kJ`
B. `1.71 xx 10^(2)kJ`
C. ` 1.81xx10^(2) kJ `
D. ` 1.91 xx 10^(2) kJ `

Explanation

To solve the problem of how much heat must be supplied to evaporate 80 grams of water, we can follow these steps:

### Step 1: Calculate the number of moles of water
We know that the molar mass of water (H₂O) is approximately 18 g/mol. To find the number of moles of 80 grams of water, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{80 \text{ g}}{18 \text{ g/mol}} \]

Calculating this gives:

\[ \text{Number of moles} = \frac{80}{18} \approx 4.44 \text{ moles} \]

### Step 2: Calculate the heat required for evaporation
The heat required to evaporate one mole of water is given as \( \Delta H_{vaporization} = 40.79 \text{ kJ/mol} \). To find the total heat required for 4.44 moles, we multiply the number of moles by the heat of vaporization:

\[ \text{Total heat} = \text{Number of moles} \times \Delta H_{vaporization} \]

Substituting the values:

\[ \text{Total heat} = 4.44 \text{ moles} \times 40.79 \text{ kJ/mol} \]

Calculating this gives:

\[ \text{Total heat} \approx 181.228 \text{ kJ} \]

### Final Answer
The heat that must be supplied to evaporate 80 grams of water is approximately **181.228 kJ**.

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