How many moles is 573.28 g of $A g C l$ $AgCl$ ?

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Answer 1 · 2016-04-09T13:37:46

$4.00 m o l A g C l$ $4.00molAgCl$

Find the relative mass of $A g C l$ $AgCl$ by adding together the relative atomic mass of each atom in the compound,

$107.868 g m o l^{- 1} A g + 35.45 g m o l^{- 1} C l = 143.318 g m o l^{- 1} A g C l$ $107.868gmol^-1Ag + 35.45gmol^-1Cl = 143.318gmol^-1AgCl$

Divide the total mass of how much $A g C l$ $AgCl$ by the mass per mole to give you the number of moles,

$\frac{573.28 g}{143.318 g m o l^{- 1}} A g C l = 4.00 m o l A g C l$ $(573.28g)/(143.318gmol^-1)AgCl = 4.00molAgCl$

How many moles is 573.28 g of AgClAgCl?