Pls inc. working! If the enthalpy change associated with freezing 3.45 g of water is 1.18kJ, calculate enthalpy change per mole of water.?

Molar mass of H₂O (water) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
No. of moles of H₂O involved in the change = (3.45 g) / (18.0 g/mol) = 0.1917 mol

Enthalpy change per mole of H₂O = (1.18 kJ) / (0.1917 mol) = 6.16 kJ/mol

Molar mass H2O = 18g/mol
Enthalpy change per mol H2O = 18/3.45*1.18 = 6.16kJ/mol