What volume of 38.6% KOH solution with a density of 1.38 g / cm3 should be taken to prepare 300 ml

What volume of 38.6% KOH solution with a density of 1.38 g / cm3 should be taken to prepare 300 ml of 0.5 M KOH solution?

Let’s find the mass of alkali required to prepare 300 ml. 0.5M potassium hydroxide solution.

N alkali = 0.3 x 0.5 = 0.15 mol;

Let’s find the mass of this amount of alkali. To do this, we multiply its molar mass by the amount of substance.

M KOH = 39 + 16 + 1 = 56 grams / mol;

The mass of alkali will be:

m KOH = 56 x 0.15 = 8.4 grams;

Let us find the volume of a 38.6% KOH solution with a density of 1.38 grams / cm3 with the same alkali content.

To do this, we divide the mass of alkali by its mass fraction and divide the resulting mass of the solution by its density.

V solution = 8.4 / 0.386 / 1.38 = 15.8 cm3 = 15.8 ml;