Question

Total no.of electrons in 81g of Al³+ are????

Answer 1

Answer:

As given Al3+

(Which means has loosed 3 e-mail )

Atomic no.=13

Electron remaining=10

27g=atomic mass

27g of Al3+=10Na

1gof Al3+=10Na/27

81gof Al3+=10*6.02*10^23*81/27=

1.806*10^25

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Answer 2

$\large{\underline{\boxed{\purple{ANSWER}}}}$

Given mass of Al³+ = 81g

Molar mass of Al³+ = 27g

So, Total no of moles :-

$no \: . \: of \: moles \: = \frac{given \: mass}{molar \: mass} \\ \\ = > no \: . \: of \: moles \: = \frac{81}{27} \\ \\ = > no \: . \: of \: moles \: = 3$

Hence, total no of atoms in 81g of Al³+ =

No. of moles × Avogadro number

= 3 × 6.022 × 10²³.

Now, total no.of electrons in Al =

13

So, total no.of electrons in Al³+ =

13 - 3 = 10

Therefore, total no.of electrons in

3 × 6.022 × 10²³ atoms of Al³+ are

==> 3 × 6.022 × 10²³ × 10

==> 3 × 6.022 × 10²⁴

==> 18.066 × 10²⁴