What volume of a 60% solution of nitric acid, the density of which is 1.305 g / ml, can be obtained

What volume of a 60% solution of nitric acid, the density of which is 1.305 g / ml, can be obtained from 150 g of nitrogen oxide 4.

Given:
ω (HNO3) = 60%
ρ solution (HNO3) = 1.305 g / ml
m (NO2) = 150 g

To find:
V solution (HNO3) -?

1) 4NO2 + O2 + 2H2O => 4HNO3;
2) M (NO2) = Mr (NO2) = Ar (N) * N (N) + Ar (O) * N (O) = 14 * 1 + 16 * 2 = 46 g / mol;
3) n (NO2) = m / M = 150/46 = 3.2609 mol;
4) n (HNO3) = n (NO2) = 3.2609 mol;
5) M (HNO3) = Mr (HNO3) = Ar (H) * N (H) + Ar (N) * N (N) + Ar (O) * N (O) = 1 * 1 + 14 * 1 + 16 * 3 = 63 g / mol;
6) m (HNO3) = n * M = 3.2609 * 63 = 205.4367 g;
7) m solution (HNO3) = m * 100% / ω = 205.4367 * 100% / 60% = 342.3945 g;
8) V solution (HNO3) = m solution / ρ solution = 342.3945 / 1.305 = 262.371 ml.

Answer: The volume of the HNO3 solution is 262.371 ml.