Solution :
1) Zinc chloride and Zinc acetate are dissociated as,
ZnCl2 = Zn2+ + 2Cl-
Zn(CH3COO)2 = Zn2+ + 2CH3COO-
Thus,
Number of moles of Zn^2+ ion is same as ZnCl2 and Zn(CH3COO)2.
>>>>Number of milimoles of Zinc chloride
= Molarity x Volume = 0.176 M x 29.3 mL = 5.1568 milimol
>>>Number of milimoles of Zinc acetate
= Molarity x Volume = 0.362 M x 11.2 mL = 4.0544 milimol
Total milimoles of Zn2+ = 5.1568 + 4 0544 = 9.2112 milimol
Total volume = 29.3 mL + 11.2 mL = 40.5 mL
Thus,
Molarity of Zn2+ ion = Total milimoles / Total volume
= 9.2112 mmol / 40.5 mL = 0.2274 M
2) Number of molecules of Zinc iodide
= 24.4 mL x 0.488 M = 11.9072 milimol
Hence, milimoles of Iodide ion = 2 x 11.9072 = 23.8144 mol
Number of molecules of Aluminium iodide
= 19.1 mL x 0.386 M = 7.3726
Hence, number of moles of Iodine ion
= 3 x 7.3726 = 22.1178 milimol
Total milimol of iodide = 23.8144 + 22.1178 = 45.9322 milimol
Total volume = 24.4 + 19.1 = 43.5 mL
Hence,
Molarity = 45.9322 mM /43.5 mL = 1.056 M
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