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I want to find out the charge on 1 gram ions of $\ce{Al^3+}$ ions.

I tried out by finding no. of moles as $ \frac{1}{27} $ , then no. of ions will be $ \frac{1}{27} × N_A $ ,

then total charge will be $ \frac{1}{27} × N_A × 3e $ but it is wrong.

The correct answer turns out to be $ N_A × 3e $ .

Where I am wrong?

[ $ e $ = charge on 1 electron, $ N_A $ = Avogadro's Number ]

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  • $\begingroup$ @RaviPrakash What you did is perfectly correct. Who told you it's wrong? $\endgroup$ – Pritt says Reinstate Monica Jul 3 '17 at 9:15
  • $\begingroup$ @PrittBalagopal Yes it seems to be correct but the answer is 3 $ N_A e $ $\endgroup$ – Fghj Jul 3 '17 at 9:23
  • $\begingroup$ They must've meant "one mole", not "one gram". Btw I didn't downvote. $\endgroup$ – Pritt says Reinstate Monica Jul 3 '17 at 9:32
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1 - gram ions is same as 1 mole.

So, we need to find only the charge on 1 mole of ions, i.e.,

$ N_A × 3e $ colulomb.

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