How many atoms of oxygen are in one MOLE of $A l_{2} {(C r O_{4})}_{3} 8 H_{2} O$ $Al_2(CrO_4)_3 8H_2O$ ?

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How many atoms of oxygen are in one MOLE of $A l_{2} {(C r O_{4})}_{3} 8 H_{2} O$ $Al_2(CrO_4)_3 8H_2O$ ?

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Answer 1 · 2016-10-24T15:14:26

Well, there are clearly (?) 20 oxygen atoms per formula unit of $A l_{2} {(C r O_{4})}_{3} \cdot 8 H_{2} O$ $Al_2(CrO_4)_3*8H_2O$ . Agreed?

Explanation:

So thus in 1 mol of the aluminum chromate hydrate there must be 20 mol of oxygen atoms, i.e. $20 \times N_{A}$ $20xxN_A$ $oxygen atoms$ $"oxygen atoms"$ , where $N_{A} = Avogadro's number$ $N_A="Avogadro's number"$ $=$ $=$ $6.022 \times 10^{23} \cdot m o l^{- 1}$ $6.022xx10^23*mol^-1$ . That's a lot of arithmetic.

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How many atoms of oxygen are in one MOLE of $A l_{2} {(C r O_{4})}_{3} 8 H_{2} O$ $Al_2(CrO_4)_3 8H_2O$ ?

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How many atoms of oxygen are in one MOLE of Al_2(CrO_4)_3 8H_2OAl2(CrO4)38H2OAl_2(CrO_4)_3 8H_2O?

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How many atoms of oxygen are in one MOLE of $A l_{2} {(C r O_{4})}_{3} 8 H_{2} O$ $Al_2(CrO_4)_3 8H_2O$ ?