Bottle (A) contains 320 mL of solution with a volume strength of 10 V. Bottle (B) contains 80 mL of — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Bottle A: $Volume Strength = 5.6 \times N \Rightarrow N = 10/5.6 = 1.785 N$. Meq of A = $320 \times 1.785 \approx 571.4$. Bottle B: Meq of B = $80 \times 5 = 400$. Total Meq = $571.4 + 400 = 971.4$. Total Volume = 400 mL. Final Normality ($N_f$) = $971.4 / 400 = 2.4285 N$. Final Volume Strength = $N_f \times 5.6 = 2.4285 \times 5.6 = 13.6 V$. Strength (g/L) = $N_f \times Eq.Wt(17) = 2.4285 \times 17 = 41.285$ g/L.