What is the molality of 0.5 molar NaNO3 solution, if density of solution is 1.2 g/ml?
Answer & explanation
Correct answer: option 1
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