The molar enthalpy of vaporization for water is 40.79 kJ/mol. Express this enthalpy of vaporization in joules per gram.
40.79 kJ/mol x (1000 J/kJ) x (1 mol/18 g) = ?
2266J/g
To express the molar enthalpy of vaporization in joules per gram, we need to convert from kJ/mol to J/g.
First, we can convert kJ to J by multiplying by 1000:
40.79 kJ/mol * 1000 J/kJ = 40790 J/mol.
Next, we need to convert moles to grams. The molar mass of water (H2O) is approximately 18.015 g/mol.
To convert from J/mol to J/g, we divide the molar enthalpy of vaporization by the molar mass of water.
40790 J/mol / 18.015 g/mol ≈ 2267.6 J/g.
Therefore, the enthalpy of vaporization for water is approximately 2267.6 J/g.
To express the molar enthalpy of vaporization for water in joules per gram, we need to convert it from kJ/mol to J/g.
First, let's convert kilojoules (kJ) to joules (J). Since 1 kJ = 1000 J, we can multiply the given value by 1000:
40.79 kJ/mol * 1000 J/1 kJ = 40,790 J/mol
Now, we need to convert moles (mol) to grams (g). The molar mass of water (H2O) is approximately 18.015 g/mol. To do the conversion, we divide the enthalpy by the molar mass:
40,790 J/mol / 18.015 g/mol ≈ 2260 J/g
Therefore, the molar enthalpy of vaporization for water is approximately 2260 J/g.