If a 1.8 molal solution of nitric acid is needed in an experiment, 25 g of the solute are available, how many grams of water must be added?

Molar mass of HNO₃ (nitric acid) = (1.0 + 14.0 + 16.0×3) g/mol = 63.0 g/mol
No. of moles of HNO₃ = (25 g) / (63.0 g/mol) = 0.397 mol

Molality = (No. of moles of HNO₃) / (Mass of water in kg)

Mass of water in kg = (No. of moles of HNO₃) / Molality = (0.397 mol) / (1.8 mol/kg) = 0.22 kg = 220 g

I'm not sure, sorry.