Density to find Molality

Question

Nitric acid, HNO₃(aq), is commercially available as a 33.7 m aqueous solution (density = 1.35 g/mL). What is the molarity of this solution?

Every time I see a problem where it gives the density to figure out the answer, my brain freezes. Please help.

Answer 1

You are being asked to essentially convert molality (m) to molarity (M). Molality means moles of HNO₃ per 1000 g of solvent, and molarity means moles of HNO₃ per 1000 ml (1 L) of solution. We are given that our solution contains 33.7 moles HNO₃ per 1000 g solvent. From this information, and the molar mass of HNO₃ of 63.0 g/mole, we can find the total mass of the solution.

Mass of solution = mass HNO₃ + mass of solvent

mass of HNO₃ = 33.7 moles x 63.0 g / mole = 2123 g HNO₃

mass of solvent (H₂O) = 1000 g

Total mass of solution = 2123 g + 1000 g = 3123 g of solution

Now, recall that molarity (M) is moles HNO₃ (33.7 mols) per LITER of solution. So, all we need to do now is to find the liters of solution. We do this by using the density provided (1.35 g/ml) and dimensional analysis.

Volume of solution = 3123 g x 1 ml / 1.35 g = 2313 mls = 2.313 liters

So, Molarity = 33.7 moles / 2.313 Liters = 14.6 M

Answer 2

Hello Evanly,

The way you approach this problem is you have to backtrack the number of moles of solute in the solution. If you have 33.7 mL, and you know the density (1.35 g/mL), you multiply the two to get 45.5 g. Then you take the 45.5 grams of Nitric Acid and by dividing the mass by the molar mass of nitric acid (63.0 g/mol), you get the 0.722 moles of nitric acid. Lastly, you take the moles and divide them by the Liter equivalent of 33.7 mL, which is 0.0337 L, to get 21.4 M of Nitric Acid, which sounds awfully concentrated.

Hope this helps,

Danil