Question

The charge on 1 gram of $A{l}^{3+}$ ions is (e=electronic charge):

• A. $\frac{1}{27}{N}_{A}ecoulomb$
• B. $\frac{1}{3}{N}_{A}ecoulomb$
• C. $\frac{1}{9}{N}_{A}ecoulomb$
• D. $3N_{A}ecoulomb$

Answer 1

The correct option is C $\frac{1}{9}{N}_{A}ecoulomb$

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1. $\frac{1}{9}$ NA e coulombs

2. So for finding the charge of one gram ion of $A{l}^{3+}$, We have to know that

1 gm atom = 1 mole of atom

So, charge on one mole of electron = $N_a\times$ charge on $1e^{⁻}$

$=6.023\times 10{²}^{23}\times 1.6\times {10}^{-19}C$.

$=9.6368\times {10}^4\left(N_a{e}^{⁻}\right)$

Thus, Charge on one mole of $A{l}^{3+}$ ion $=3\times 96368$

$=289104C\left(or3N_a{e}^{⁻}\right)$

Since , 27 g of $A{l}^{3+}$ ion having charge of 289104 C .

Therefore, 1 g of $A{l}^{3+}$ ion having charge = $\frac{289104}{27}$ C.

$=10707.5C.\left(or\frac{N_a{e}^{⁻}}{9}\right)$