# Latent heat of vaporization

The molecular weight of water is 18.015 gram. The number of moles of water in one liter (1000 gram) will be: $3.34\times 10^{25}$ molecules (in 1kg). We know that latent heat of vaporization of water is $L_v= 2.26\times 10^6$ then the amount of heat required to rise one molecule of water will be $1.47\times 10^{-19}$ joule: Does this calculation is correct according to macroscopic and microscopic physics?

$1$ mole of water = $18.015$ g so $1$ kg water = $1000/18.015 = 55.51$ moles.
$1$ kg of water = $55.51 \times 6.023 \times 10^{23} = 3.34 \times 10^{25}$ molecules.
Latent heat of water per kg = $2.257 \times 10^6$ J.
Latent heat per molecule = $2.257 \times 10^6 / 3.34 \times 10^{25} = 6.76 \times 10^{-20}$J.