What volumes of solutions with a mass fraction of sodium hydroxide of 32% (density 1.35 g / ml)

What volumes of solutions with a mass fraction of sodium hydroxide of 32% (density 1.35 g / ml) and with a mass fraction of phosphoric acid 46% (density 1.3 g / ml) will be required to obtain sodium dihydrogen phosphate with a mass of 24 g.

Given:
ω (NaOH) = 32%
ρ solution (NaOH) = 1.35 g / ml
ω (H3PO4) = 46%
ρ solution (H3PO4) = 1.3 g / ml
m (NaH2PO4) = 24 g

To find:
V solution (NaOH) -?
V solution (H3PO4) -?

1) NaOH + H3PO4 => NaH2PO4 + H2O;
2) n (NaH2PO4) = m / M = 24/120 = 0.2 mol;
3) n (NaOH) = n (NaH2PO4) = 0.2 mol;
4) m (NaOH) = n * M = 0.2 * 40 = 8 g;
5) m solution (NaOH) = m * 100% / ω = 8 * 100% / 32% = 25 g;
6) V solution (NaOH) = m solution / ρ solution = 25 / 1.35 = 18.52 ml;
7) n (H3PO4) = n (NaH2PO4) = 0.2 mol;
8) m (H3PO4) = n * M = 0.2 * 98 = 19.6 g;
9) m solution (H3PO4) = m * 100% / ω = 19.6 * 100% / 46% = 42.609 g;
10) V solution (H3PO4) = m solution / ρ solution = 42.609 / 1.3 = 32.78 ml.

Answer: The volume of the NaOH solution is 18.52 ml; H3PO4 – 32.78 ml.