Sodium nitrate NaNO3 was mixed in an amount of 0.1 mol with 0.5 l of water what

Sodium nitrate NaNO3 was mixed in an amount of 0.1 mol with 0.5 l of water what is the mass fraction of sodium hydroxide in the resulting solution.

Given:
n (NaNO3) = 0.1 mol
V (H2O) = 0.5 l = 500 ml

Find:
ω (NaNO3) -?

Solution:
1) M (NaNO3) = Mr (NaNO3) = Ar (Na) * N (Na) + Ar (N) * N (N) + Ar (O) * N (O) = 23 * 1 + 14 * 1 + 16 * 3 = 85 g / mol;
2) m (NaNO3) = n (NaNO3) * M (NaNO3) = 0.1 * 85 = 8.5 g;
3) m (H2O) = ρ (H2O) * V (H2O) = 1 * 500 = 500 g;
4) m solution (NaNO3) = m (NaNO3) + m (H2O) = 8.5 + 500 = 508.5 g;
5) ω (NaNO3) = m (NaNO3) * 100% / m solution (NaNO3) = 8.5 * 100% / 508.5 = 1.7%.

Answer: The mass fraction of NaNO3 is 1.7%.