How many grams of 60% KOH is required to neutralize 225 ml of a 35% (p = 1.2 g / ml) nitric acid solution.

Given:
ω (KOH) = 60%
V solution (HNO3) = 225 ml
ω (HNO3) = 35%
ρ solution (HNO3) = 1.2 g / ml

To find:
m solution (KOH) -?

Decision:
1) KOH + HNO3 => KNO3 + H2O;
2) m solution (HNO3) = ρ solution (HNO3) * V solution (HNO3) = 1.2 * 225 = 270 g;
3) m (HNO3) = ω (HNO3) * m solution (HNO3) / 100% = 35% * 270/100% = 94.5 g;
4) n (HNO3) = m (HNO3) / M (HNO3) = 94.5 / 63 = 1.5 mol;
5) n (KOH) = n (HNO3) = 1.5 mol;
6) m (KOH) = n (KOH) * M (KOH) = 1.5 * 40 = 60 g;
7) m solution (KOH) = m (KOH) * 100% / ω (KOH) = 60 * 100% / 60% = 100 g.

Answer: The mass of the KOH solution is 100 g.