How much energy does it take to boil 100 mL of water? (Refer to table of constants for water.)A.100 mL × × × 6.03 kJ/mol = 33.5 kJB.100 mL × × × (–285.83 kJ)/mol = –1586 kJC.100 mL × × × 4.186 kJ/mol = 23.2 kJD.100 mL × × × 40.65 kJ/mol = 226 kJ

To answer this question, we need to know the specific heat capacity of water and the heat of vaporization. The specific heat capacity of water is approximately 4.186 J/g°C and the heat of vaporization is approximately 40.7 kJ/mol.

First, we need to convert 100 mL of water to grams. Since the density of water is approximately 1 g/mL, 100 mL of water is approximately 100 grams.

Next, we need to calculate the amount of energy required to heat the water from room temperature (assume 25°C) to the boiling point of water (100°C). Using the formula q = mcΔT, where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature:

q = (100 g)(4.186 J/g°C)(100°C - 25°C) = 31,395 J or 31.4 kJ

Then, we need to calculate the amount of energy required to change the water from a liquid at its boiling point to a gas. This is where we use the heat of vaporization. However, the heat of vaporization is given per mole, so we need to convert the mass of water to moles. Since the molar mass of water is approximately 18 g/mol:

100 g ÷ 18 g/mol = 5.56 mol

Then, the energy required to vaporize the water is:

q = (5.56 mol)(40.7 kJ/mol) = 226.3 kJ

Finally, we add the two amounts of energy together:

31.4 kJ + 226.3 kJ = 257.7 kJ

So, it takes approximately 258 kJ of energy to boil 100 mL of water. Therefore, none of the given options (A, B, C, D) are correct.