Question

Calculate the molalitey of a 17.5% (by mass) aqueous soltuon of nitric acid.

Answer 1

Solution :-

Lets assume we have 1 L solution density = 1 g/ml

then mass of solution = 1000 g

so the mass of nitric acid

1000 g * 17.5 % /100 % = 175 g

therefore mass of water = 1000 g - 175 g = 825 g

niow lets calculate moles of the HNO3

moles = mass / molar mass

moles of HNO3 = 175 g / 63.01 g per mol = 2.78 mol HNO3

molality = moles / kg solvent

molality = 2.78 mol / 0.825 kg

            = 3.37 m

Therefore molality of the solution is 3.37 m