To determine the amount of heat required to convert one mole of liquid water to vapor at [tex]\(100^{\circ} C\)[/tex], given the heat of vaporization, we need to follow a few steps. Let's break it down.
### Step 1: Understand the Heat of Vaporization
The heat of vaporization is the amount of energy required to transform a given quantity of a substance from a liquid into a gas at a specific temperature. For water, the heat of vaporization is given as [tex]\(2260.87 \, \text{J/g}\)[/tex].
### Step 2: Identify the Molar Mass of Water
The molar mass of water (H₂O) is the sum of the masses of its constituent atoms:
- Hydrogen has a molar mass of approximately [tex]\(1.008 \, \text{g/mol}\)[/tex].
- Oxygen has a molar mass of approximately [tex]\(16.00 \, \text{g/mol}\)[/tex].
Thus, the molar mass of water is:
[tex]\[
\text{Molar mass of H₂O} = 2 \times 1.008 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.015 \, \text{g/mol}
\][/tex]
### Step 3: Calculate the Heat Required for One Mole
To find the total heat required to vaporize one mole of water, we multiply the molar mass of water by the heat of vaporization per gram:
[tex]\[
\text{Heat required per mole} = \text{Heat of vaporization} \times \text{Molar mass of water}
\][/tex]
[tex]\[
\text{Heat required per mole} = 2260.87 \, \text{J/g} \times 18.015 \, \text{g/mol}
\][/tex]
### Step 4: Perform the Calculation
Now let's multiply these values:
[tex]\[
\text{Heat required per mole} = 2260.87 \, \text{J/g} \times 18.015 \, \text{g/mol} = 40729.57305 \, \text{J/mol}
\][/tex]
### Step 5: Round to Two Decimal Places
We need to express the answer rounded to two decimal places:
[tex]\[
40729.57305 \, \text{J/mol} \approx 40729.57 \, \text{J/mol}
\][/tex]
### Final Answer
The amount of heat required to convert one mole of liquid water to vapor at [tex]\(100^{\circ} C\)[/tex] is:
[tex]\[
40729.57 \, \text{J/mol}
\][/tex]
