124 gm of mixture containing NaHCO 3 , AlCl 3 \& KNO 3 requires 500 ml, 8% w/w NaOH solution [ d NaOH = 1.8 gm/ml] for complete neutralisation. On heating same amount of mixture, it shows loss

Question

124 gm of mixture containing NaHCO 3 ​ , AlCl 3 ​ \& KNO 3 ​ requires 500 ml, 8% w/w NaOH solution [ d NaOH ​ = 1.8 gm/ml] for complete neutralisation. On heating same amount of mixture, it shows loss in weight of 18.6 gm. Calculate mole percent of KNO 3 ​ in the original mixture. Weak base formed doesn't interfere in reaction. Assume KNO 3 ​ does not decompose under given conditions.

TestHub · Accepted Answer

Step 1 (NaOH Moles): Total mass of solution = $500 m l \times 1.8 g / m l = 900 g$ ; moles of NaOH = $\frac{900 \times 0.08}{40} = 1.8 m o l es$ .

Step 2 (NaHCO₃ Moles): Heating $2 m o l es$ of NaHCO₃ loses $62 g$ (H₂O + CO₂); thus, moles of NaHCO₃ = $\frac{18.6 g}{31 g / m o l} = 0.6 m o l es$ (mass = $50.4 g$ ).

Step 3 (AlCl₃ Moles): Neutralization requires $n_{N a O H} = n_{N a H C O_{3}} + 3 n_{A lC l_{3}} \to 1.8 = 0.6 + 3 n_{A lC l_{3}}$ , giving $0.4 m o l es$ of AlCl₃ (mass = $53.4 g$ ).

Step 4 (KNO₃ Moles): Mass of KNO₃ = $124 - (50.4 + 53.4) = 20.2 g$ ; moles of KNO₃ = $\frac{20.2}{101} = 0.2 m o l es$ .

Step 5 (Mole %): Total moles = $0.6 + 0.4 + 0.2 = 1.2$ ; mole $%$ of KNO₃ = $(\frac{0.2}{1.2}) \times 100 = 16.67%$ .

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