10 ml of (N/2) HCl, 30 ml of (N/10) HNO₃ and 75 ml of (N/5) HNO₃ are mixed, the normality of H⁺ in the resulting solution is

The equiv. of H⁺ in 30 ml of (N / 2) HCl = (10 / 2) 10^–3 The equiv. of H⁺ in 30 ml of (N / 10) HNO3 = (30 / 10) * 10^–3 The equiv. of H⁺ in 75 ml of (N / 10) HNO3 = (75 / 5) * 10^–3. Hence, total equiv. of H⁺ (5 + 3 + 15) * 10^–3 = 23 10^–3 Total volume of solution = 115 ml Hence, normality of H⁺ in the resulting mixture = [(23 * 10^–3 * 10^3) / 115) (N) = (N / 5) = 0.2 (N). Hence, this is a correct option.