Given, molarity =2M [i.e 2 mole NaOH in 1 L solution] Density (d)=1.28 cm−3 Molecular weight of NaOH=40gmol−1 We know that, Density = volume of solution mass of solution ∴ Mass of solution =1.28 g cm−3×1000 mL=1280 g Moreover, molarity = volume of solution ( in mL) number of moles of solute ×1000∴2=40×1000 mass of solute ×1000=80 g Now, mass of solution = mass of solvent + mass of solute 1280 g= mass of solvent +80∴ Mass of solvent =1280−80=1200 g=1.2 kg Now, molality = mass of solvent ( in kg) number of moles of solute =1.22=1220=35=1.67 m Alternative method Molality (m)=1000 d−MωtM×1000 where, M= molarity, d= density of solution Mωt= molar mass of solute. On substing the given values, we get Molality =1000×1.28−402×1000=1280−402×1000=12402000=1.612 m