Question

Number of electrons present in 9.6 grams of so4 -2 ion

Answer 1

Total e in SO4-2 ion=50e

moles in 9.6 gm of SO4-2=1/10

(taking M.W of so4-2 approx 96 gm.)

so,mole =given no.of molicele/Na(6*10 23)

so mole =6 into 10 to the power 22

so no of e in 9.6 gm of SO4-2 =3* 10 to the power 24e

Answer 2

The number of electrons present in the given amount of sulfate ion are $2.05\times 10^{24}$

Explanation:

The chemical formula of sulfate ion is $SO_4^{2-}$

Number of electrons in sulfur atom = $1s^22s^22p^63s^33p^4$ = 16

Number of electrons in oxygen atom = $1s^22s^22p^4$ = 8

Charge on the ion = -2

If the charge is negative, the charge will be added up in the total number of electrons and is the charge is positive, the charge will be subtracted from the total number of electrons

Total number of electrons in $SO_4^{2-}=[16+(2\times 8)+2]=34$

To calculate the number of moles, we use the equation:

Given mass of sulfate ion = 9.6 g

Molar mass of sulfate ion = 96 g/mol

Putting values in above equation, we get:

$\text{Moles of }SO_4^{2-}=\frac{9.6g}{96g/mol}=0.1mol$

According to mole concept:

1 mole of a substance contains $6.022\times 10^{23}$ number of particles

So, 0.1 moles of sulfate ion will contain $(34\times 0.1\times 6.022\times 10^{23})=2.05\times 10^{24}$ number of electrons

Learn more about mole concept:

https://brainly.in/question/4988828

https://brainly.in/question/6634081

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