The amount of energy required to change 20g of water to steam is +50kj Mol calculate the standard heat of vaporization
I'm confused by your post. Do you mean 50 kJ/mol or do you mean 50 kJ for 20 g.
50 kj/mol
To calculate the standard heat of vaporization (ΔHvap), we need to use the equation:
ΔHvap = q / n
Where:
ΔHvap = standard heat of vaporization
q = heat energy required (in joules)
n = moles of substance
Given:
q = +50 kJ (converted to joules by multiplying by 1000) = 50,000 J
m = 20 g
Molar mass of water (H2O) = 18 g/mol
First, we calculate the number of moles (n) using the molar mass:
n = m / M
n = 20 g / 18 g/mol
n ≈ 1.11 mol
Now, we can plug the values into the equation to calculate ΔHvap:
ΔHvap = 50,000 J / 1.11 mol
ΔHvap ≈ +45,045 J/mol
Therefore, the standard heat of vaporization for water is approximately +45.045 kJ/mol.
To calculate the standard heat of vaporization, you need to know the molar mass of water and the number of moles.
1. Determine the molar mass of water (H2O).
The molecular weight of hydrogen (H) is 1 g/mol, and the molecular weight of oxygen (O) is 16 g/mol. Therefore, the molar mass of water is (2 * 1 g/mol) + (1 * 16 g/mol) = 18 g/mol.
2. Convert the mass of water given (20g) to moles.
Moles = mass / molar mass
Moles = 20g / 18 g/mol ≈ 1.11 moles (rounded to two decimal places).
3. Calculate the heat of vaporization.
Heat of vaporization = energy / moles
Heat of vaporization = 50 kJ / 1.11 mol ≈ 45 kJ/mol (rounded to two decimal places).
Therefore, the standard heat of vaporization of water is approximately 45 kJ/mol.