Answer to Question #319944 in General Chemistry for jjoel

Question

What is the molality of a solution composed of 40.0 g sodium nitrate (NaNO3) and 500.0 g of water? 

Answer 1

Solution:

Molality = Moles of solute / Kilograms of solvent

solute = NaNO₃

solvent = H₂O (water)

The molar mass of NaNO₃ is 84.995 g/mol

Therefore,

Moles of NaNO₃ = (40.0 g NaNO₃) × (1 mol NaNO₃ / 84.995 g NaNO₃) = 0.4706 mol NaNO₃

Moles of NaNO₃ = 0.4706 mol

Kilograms of water = (500.0 g H₂O) × (1 kg / 1000 g) = 0.5 kg H₂O

Thus:

Molality of NaNO₃ solution = Moles of NaNO₃ / Kilograms of water

Molality of NaNO₃ solution = (0.4706 mol) / (0.5 kg) = 0.94 mol/kg = 0.94 m

Molality of NaNO₃ solution = 0.94 m

Answer: The molality of NaNO₃ solution is 0.94 m