Question

How many moles and number of ions of each type are present in the following aqueous solution?

356mL of 0.308g aluminum sulfate/L

__________mol of aluminum

______ x 10^ _______ aluminum ions

__________mol of sulfate

______ x 10^ _______ sulfate ions

Answer 1

volume = 356 ml = 0.356 L

concentration of Al2(SO4)3 = 0.308 g/L

molar mass of Al2(SO4)3 = 342.15 g/mol

molarity of aluminum sulfate = 0.308 / 342.15 = 9 x 10^-4 M

moles = molarity x volume = 9 x 10^-4 x 0.356 = 3.2 x 10^-4

moles of aluminum sulfate = 3.2 x 10^-4

moles of aluminium = 2 x 3.2 x 10^-4 = 6.41 x 10^-4

moles of aluminium = 6.41 x 10^-4

aluminum ions = 6.41 x 10^-4 x 6.023 x 10^23 = 3.86 x 10^20

aluminum ions = 3.86 x 10^20

moles of sulfate = 3 x 3.2 x 10^-4 = 9.6 x 10^-4

moles of sulfate9.6 x 10^-4

sulfate ions = 9.6 x 10^-4 x 6.023 x 10^23 = 5.78 x 10^20

sulfate ions = 5.78 x 10^20