How much water at its boiling point can be vaporized by adding 1.50 kJ of heat? ΔHvap for H2O = 40.7 kJ/mol
To calculate the amount of water that can be vaporized, we need to use the molar enthalpy of vaporization (ΔHvap) for water.
Given:
Heat added (q) = 1.50 kJ
ΔHvap = 40.7 kJ/mol
We can use the following equation to find the amount of water vaporized:
q = n * ΔHvap
Where:
q = heat added in joules
n = number of moles of water vaporized
ΔHvap = molar enthalpy of vaporization
First, we need to convert the heat from kilojoules (kJ) to joules (J):
1 kJ = 1000 J
Therefore,
q = 1.50 kJ * 1000 J/kJ
q = 1500 J
Next, we can rearrange the equation to solve for n:
n = q / ΔHvap
n = 1500 J / 40.7 kJ/mol
Now, let's convert kJ to J and solve for n:
n = 1500 J / (40.7 kJ/mol * 1000 J/kJ)
n = 0.0369 mol
Therefore, by adding 1.50 kJ of heat, approximately 0.0369 moles of water can be vaporized at its boiling point.
To determine the amount of water that can be vaporized by adding 1.50 kJ of heat, we need to consider the heat required for vaporization of water (ΔHvap) and the molar mass of water (H2O).
The equation that relates the amount of heat (q), the heat of vaporization (ΔHvap), and the number of moles (n) of a substance is:
q = ΔHvap * n
We can rearrange this equation to solve for the number of moles:
n = q / ΔHvap
Given that the heat added is 1.50 kJ and the heat of vaporization for water is 40.7 kJ/mol, we can substitute these values into the equation:
n = (1.50 kJ) / (40.7 kJ/mol)
Next, we need to convert kilojoules (kJ) to joules (J) since the molar heat of vaporization is given in joules per mole (J/mol).
1 kJ = 1000 J
Therefore, 1.50 kJ is equal to 1500 J.
Now, we can substitute the values into the equation:
n = (1500 J) / (40.7 kJ/mol)
Now we can calculate the number of moles of water vaporized.
40.7 kJ/mol x #mols = kJ
40.7 kJ/mol x #mols = 1.50 kJ.
Solve for # mols. If you want grams, convert mols to grams by
grams = mols x molar mass = ?