Question

How would you determine the molality of a solution that is 45.0% (m/m) NaNO3(aq), and has a density of 1.37 g/mL? I'm a bit confused. An explanation with work provided would be super helpful, thank you.

Answer 1

m/m percentage means mass of solute(in g) present per 100 g of mixture.

45% m/m NaNO_​​​3 means 45 g NaNO_​​​3​​ is present in 100 g mixture or solution.

So, 1000 g solution will contain 450 g NaNO_​​​3 .

Molar mass of NaNO_​​​3 = 85 g/mol.

We known, no of moles = (given mass/molar mass)

So, 450 g NaNO_​​​3 = (450/85)= 5.3 mole NaNO_​​​3 .

And mass of solvent = (450-85)g = 365 g

Molality is the moles of solute present per kg (1000g ) of solvent.

Here, 365 g solution contains 5.3 moles NaNO_​​​3 .

So, 1000 g solution will contain (5.3/365)×1000 = 14.5 moles

So, molality of the solution is 14.5 m.