4.17) How many total moles of ions are released when each ofthe following sample
ID: 687469 • Letter: 4
Question
4.17) How many total moles of ions are released when each ofthe following samples dissolves completely in water? a) .734 moles of NaPO4 b)3.86 g of CuSO4*5H2O c)8.66*10^20 formula units of NiCl2 I know how to convert grams and formula units into moles, buti'm stuck on how to find the number of moles that are relaesed whenthe substance is disolved in water. Any help would be muchappreciated 4.17) How many total moles of ions are released when each ofthe following samples dissolves completely in water? a) .734 moles of NaPO4 b)3.86 g of CuSO4*5H2O c)8.66*10^20 formula units of NiCl2 I know how to convert grams and formula units into moles, buti'm stuck on how to find the number of moles that are relaesed whenthe substance is disolved in water. Any help would be muchappreciatedExplanation / Answer
a) In 0.734 moles of Na3PO4 thatis dissociated in to 3*0.734 moles of Na+ nad 1*0.734moles of PO4-3 . total moles of ions in thesolution = 3*0.734 + 1*0.734 = 2.936 moles of ions arepresent b) The number of moles of CuSO4*5H2O present in thesolution = 3.86 g/ 249.685 g/mol = 0.0154 mole In solution this hydrated salt dissolved and forms1*0.0154 mole Cu+2 and 1 mole ofSO4-2 hence - total number of moles of ions present = 0.0154 mol +0.0154 mol = 0.0308 moles of ions c) 8.66*1020 formula units ofNiCl2 means 8.66*1020 Ni+2 and2*8.66*1020 Cl- Therefore number of moles of Ni+2 =8.66*1020 / 6.023*1023 = 1.437*10-3 mole Number of moles of Cl- =2*8.66*1020 Cl- / 6.023*1023 = 2.87*10-3 mole Therfore total moles of ions in the solution =1.437*10-3 mole + 2.87*10-3 mole = 4.307*10-3 mole = 2.87*10-3 mole Therfore total moles of ions in the solution =1.437*10-3 mole + 2.87*10-3 mole = 4.307*10-3 moleRelated Questions
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