Answer:
The heat required to evaporate 228.3 g of water at 100.0°C is approximately 5.16 × 10^5 J.
Step-by-step explanation:
To calculate the heat required to evaporate a given amount of water, we can use the formula:
Heat = mass × molar heat of vaporization
Given:
Mass of water (m) = 228.3 g
Molar heat of vaporization (ΔH) = 4.07 × 10^4 J/mol
First, we need to convert the mass of water from grams to moles. The molar mass of water (H₂O) is approximately 18.015 g/mol.
Moles of water (n) = mass (m) / molar mass (M)
= 228.3 g / 18.015 g/mol
≈ 12.67 mol
Now we can calculate the heat required using the formula:
Heat = mass × molar heat of vaporization
= n × ΔH
Heat = 12.67 mol × 4.07 × 10^4 J/mol
≈ 5.16 × 10^5 J