How much heat is necessary to completely vapourize 75.0 g of ice at 0°C?

Molar mass of H₂O (ice/water/water vapor) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
No. of moles of 75.0 g of H₂O = (75.0 g) / (18.0 g/mol) = 75.0/18.0 mol

Amount of heat needed to melt 75.0 g of ice at 0°C
= (75.0/18.0 mol) × (6.02 kJ/mol)
= (75.0/18.0) × 6.02 kJ

Amount of heat needed to heat up 75.0 g of water from 0°C to 100°C
= (75.0 g) × (4.18 J/g °C) × (100 - 0)°C
= 75.0× (418/1000) × 100 kJ

Amount of heat needed to vaporizae 75.0 g of water to vapor at 100°C
= (75.0/18.0 mol) × (40.6 kJ/mol)
= (75.0/18.0) × 40.6 kJ

Amount of heat needed to vaporize 75.0 g of ice at 0°C
= [(75.0/18.0) × 6.02] + [75.0× (4.18/1000) × 100] + [(75.0/18.0) × 40.6] kJ
= 225.6 kJ