How many grams of 5% potassium chloride solution is required to react with 80 ml of 0.5N silver nitrate solution?

Given:
ω (KCl) = 5%
V (HNO3) = 80 ml = 0.08 l
SN (HNO3) = 0.5 n

To find:
m solution (KCl) -?

Decision:
1) KCl + HNO3 => KNO3 + H2O;
2) n (HNO3) = CH (HNO3) * V solution (HNO3) / z = 0.5 * 0.08 / 1 = 0.04 mol;
3) n (KCl) = n (HNO3) = 0.04 mol;
4) M (KCl) = Mr (KCl) = Ar (K) * N (K) + Ar (Cl) * N (Cl) = 39 * 1 + 35.5 * 1 = 74.5 g / mol;
5) m (KCl) = n (KCl) * M (KCl) = 0.04 * 74.5 = 2.98 g;
6) m solution (KCl) = m (KCl) * 100% / ω (KCl) = 2.98 * 100% / 5% = 59.6 g.

Answer: The mass of the KCl solution is 59.6 g.