How many gram-ions of $S O_{4}^{2 -}$ are present in 2 gram-molecules of $F e S O_{4} . (N H_{4})_{2} S O_{4} .6 H_{2} O$ ?

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Solution

$2 gram molecules of F e S O_{4} . (N H_{4})_{2} S O_{4} .6 H_{2} O = 2 moles of F e S O_{4} . (N H_{4})_{2} S O_{4} .6 H_{2} O$

From the formula unit,
1 mole of $F e S O_{4} . (N H_{2})_{2} S O_{4} .6 H_{2} O$ has 2 mole of $S O_{4}^{2 -}$ .

So, 2 mole of $F e S O_{4} . (N H_{4})_{2} S O_{4} .6 H_{2} O$ has 4 moles of $S O_{4}^{2 -}$ ions.

Hence, 4 gram ions of $S O_{4}^{2 -}$ are present in 2 gram molecule of $F e S O_{4} . (N H_{4})_{2} S O_{4} .6 H_{2} O$ .

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