# How to calculate the charge on 1 gram ions Aluminum cation?

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 ]

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

$N_A × 3e$ colulomb.