How many grams of 5% NaOH solution is required to neutralize 50 g of 3% nitric acid?

Given:
ω (NaOH) = 5%
m solution (HNO3) = 50 g
ω (HNO3) = 3%

To find:
m solution (NaOH) -?

Decision:
1) NaOH + HNO3 => NaNO3 + H2O;
2) m (HNO3) = ω (HNO3) * m solution (HNO3) / 100% = 3% * 50/100% = 1.5 g;
3) M (HNO3) = 63 g / mol;
4) n (HNO3) = m (HNO3) / M (HNO3) = 1.5 / 63 = 0.02 mol;
5) n (NaOH) = n (HNO3) = 0.02 mol;
6) M (NaOH) = 40 g / mol;
7) m (NaOH) = n (NaOH) * M (NaOH) = 0.02 * 40 = 0.8 g;
8) m solution (NaOH) = m (NaOH) * 100% / ω (NaOH) = 0.8 * 100% / 5% = 16 g.

Answer: The mass of the NaOH solution is 16 g.