What is the molality of 0.5 molar NaNO₃ solution, if density of solution is 1.2 g/ml?

0.433

0.5 M NaNO₃ solution = 0.5 moles (42.5 gm) of NaNO₃ in 1 litre solution 

Mass of solvent = Mass of solution - Mass of solute

                       = (Volume of solution x density) - Mass of solute

                       = (1000 x 1.2) - 42.5

                       = 1200 - 42.5

                       = 1157.5 g

Molality, m = \(\frac{0.5}{1157.5}\)x1000 = 0.433