How many grams nano3 add to 500g of h20 prepare solution .5 molal nano3?

Answer is: mass of sodium nitrate is 21,25 g.
m(H₂O) = 500 g · 1 kg/1000 g = 0,5 kg.
b(solution) = 0,500 m = 0,500 mol/kg.
m(NaNO₃) = ?
b(solution) = n(NaNO₃) ÷ m(H₂O).
n(NaNO₃) = 0,500 mol/kg · 0,5 kg.
n(NaNO₃) = 0,25 mol.
m(NaNO₃) = n(NaNO₃) · M(NaNO₃).
m(NaNO₃) = 0,25 mol · 85 g/mol.
m(NaNO₃) = 21,25 g.

Particle exchange energy though elastic collision , hope this .