How many grams of technical caustic soda containing 97.00% NaOH will be required

How many grams of technical caustic soda containing 97.00% NaOH will be required to prepare 1500 g of 10.00% NaOH solution?

Let’s calculate the weight of the alkali.

M alkali = 1500 x 0.1 = 150 grams;

Taking into account the alkali content of 97%, it is necessary to hang:

M weight of alkali = 150 / 0.97 = 154.64 grams;

In the same way, we find the weight of the solvent (water).

(In the calculations, we take the density of water equal to 1 gram / cm3).

M water (solvent) = 1500 x 0.9 = 1350 grams = 1350 ml;

Let’s check. To do this, add the masses of the components of the solution and get the weight of the solution in question.

150 + 1350 = 1500;