Explanation
So for finding the charge of one gram ion of $$Al^{3+}$$, We have to know that
1 gm atom = 1 mole of atom
So, charge on one mole of electron = $$N_a \times$$ charge on $$1 e^⁻$$
Thus, Charge on one mole of $$Al^{3+}$$ ion $$= 3 \times N_A \times e$$
Since , 27 g of $$Al^{3+}$$ ion having charge of $$ 3 \times N_A \times e$$.
Therefore, 1 g of $$Al^{3+}$$ ion having charge = $$\cfrac{3 \times N_A \times e}{ 27}$$ C.
$$=\cfrac{ N_a e^⁻}{ 9} C$$
Option C is correct.
Nuclear radius $$=10^{-15}$$ m
Atomic radius $$= 10^{-10}$$ m
Ratio $$= \dfrac{10^{-15}}{ 10^{-10}}=10^{-5}$$
Hence, the correct option is $$C$$
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