Evaporation of a 160 g NaNO3 solution resulted in a dry residue weighing 4 g. What is the mass fraction

Evaporation of a 160 g NaNO3 solution resulted in a dry residue weighing 4 g. What is the mass fraction of salt in the solution?

The residue after evaporation is the mass of the solute:
m (r.v. NaNO3) = w * m (solution), where w is the mass fraction of the solute NaNO3, and m (solution) is the mass of the solution.
Find w = m (r.v. NaNO3) / m (r-ra).
w = 4/160 = 0.025 or 2.5%
If the residue was overheated during drying, and then collected in an airtight container, then a reaction could take place
2NaNO3 = 2NaNO2 + O2, and then the mass of the residue m (rest) = 4 g refers to NaNO2.
In this case, we find its amount of substance n (NaNO2) = m (rest) / M (NaNO2), where M (NaNO2) = 69 g / mol is the molar mass of sodium nitrite
n (NaNO2) = 4/69 = 0.058 mol
The reaction equation shows that n (NaNO3) = n (NaNO2) = 0.058 mol.
Then the mass of the original sodium nitrate:
m (NaNO3) = n (NaNO3) * M (NaNO3) = 0.058 * 85 = 4.93 g,
M (NaNO3) = 85 g / mol – molar mass of sodium nitrate
w = 4.93 / 160 = 0.031 or 3.1%