# Technical potassium hydroxide weighing 3.5 g was dissolved in water, receiving 500 ml of solution. According to chemical analysis

**Technical potassium hydroxide weighing 3.5 g was dissolved in water, receiving 500 ml of solution. According to chemical analysis, the molar concentration of the solution was 0.1 mol / L. What is the mass fraction of KOH in the substance**

Given:

Mr (KOH) = 56g / mol;

Cm (KOH) = n (KOH) = 0.1 mol / L;

V (r-ra) = 500ml or 0.5l;

mtechn. (KON) = 3.54g.

w% (KOH) =?

Decision:

First, calculate the amount of potassium hydroxide in the solution:

n (KOH) = Cm * V (solution) = 0.1 mol / L * 0.5 L = 0.05 mol.

Find the mass of pure potassium hydroxide:

m (KOH) = n (KOH) * M (KOH) = 0.05 mol * 56 g / mol = 2.8 g

Now we calculate the mass fraction of KOH in the technical sample of it, we get:

w% (KOH) = m (KOH) * 100% / mtechn. (KOH) = 2.8 g * 100% / 3.5 g = 80% or 0.8