Find the mass of sodium nitrate required to prepare one liter of a solution with a mass fraction ω (NaNO3) equal to 35%

Find the mass of sodium nitrate required to prepare one liter of a solution with a mass fraction ω (NaNO3) equal to 35% (density (P solution) is 1.27 g / cm3). Can you explain how to solve this?

Let’s calculate the weight of the dissolved salt.

M salt = 1000 x 1.27 x 0.35 = 444.5 grams;

Similarly, we find the weight of the solvent (water).

(In the calculations, we will make the assumption that the density of water is 1 gram / ml).

M water = 1000 x 1.27 x 0.65 = 825.5 grams = 825.5 ml;

Let’s check. To do this, add the masses of the components of the solution and get the mass of the required solution.

444.5 + 825.5 = 1270;