Question

Calculate the total number of electrons present in 48 grams of mg2+

Answer 1

20×6.02×10^23
24g Mg is 1 mol so 48g Mg is 2 mol Mg2+ has lost 2 electrons so has total now 10×NA for two mol it becomes 20×NA

Answer 2

Answer: $120.46\times 10^{23}$ electrons

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}$

For $Mg^{2+}$

Mass given = 48 g

Molar mass of  $Mg^{2+}$ = 24 g/mol

Putting values in above equation, we get:

$\text{Moles of}Mg^{2+}=\frac{48g}{24g/mol}=2mol$

$Mg:12:1s^22s^22p^63s^2$

$Mg^{2+}:10:1s^22s^22p^6$

According to Avogadro's law, 1 mole of every substance contains avogadro's number $(6.023\times 10^{23})$ of particles.

1 mole of $Mg^{2+}$ will contain=$10\times {6.023\times 10^{23}=60.23\times 10^{23}$ electrons

2 moles of $Mg^{2+}$ will contain=$\frac{60.23\times 10^{23}}{1}\times 2=120.46\times 10^{23}$ electrons