Question

How many total atoms are contained in 75 g of arsenic (V) arsenate? How many total ions are contained in 75 g of arsenic (V) arsenate?

Answer 1

weight of arsenic (V) arsenate = 75g

gram molecular weight of arsenic (V) arsenate = 919.36g/mole

no of moles of arsenic (V) arsenate = 75/919.36   = 0.0816moles

atomicity of As3(Aso4)5 is 28

total no of atoms arsenic(v) arsent   = no of moles *6.023*10^23 *atomicity

                                                           = 0.0816*6.023*10^23 *28   = 1.376*10^24 atoms

As3(AsO4)5 -------------> 3As^5+ (aq) + 5AsO4^3- (aq)

                                       3 ions             5 ions

8 ions is As3(AsO4)5

Total no of ions in As3(AsO4)5   = no of moles * 6.023*10^23 * ions

                                                   = 0.0816*6.023*10^23 * 8   = 3.93*10^23 ions >>>answer