Question

Find the charge of $48g$ of ${Mg}^{2+}$ ions in coulombs.

• A. $2.4\times {10}^{23}C$
• B. $6.82\times {10}^{5}C$
• C. $3.86\times {10}^{5}C$
• D. $1.93\times {10}^{5}C$

Answer 1

The correct option is C $3.86\times {10}^{5}C$

Molar mass of ${Mg}^{+2}=24g$

Given mass of ${Mg}^{+2}=48g$

No. of moles of ${Mg}^{+2}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{48}{24}=2\text{moles}$

1 mole of ${Mg}^{+2}$ = 2 moles of electrons

$\therefore$2 moles of ${Mg}^{+2}$ = 4 moles of electrons

$\therefore$ No. of electrons = no. of moles of electrons $\times$ Avogadro's constant

$\Rightarrow$ No. of electrons $=4\times 6.023\times {10}^{23}=2.4\times {10}^{24}$

As we know that,

Charge on 1 electron $=1.6\times {10}^{-19}C$

$\therefore$ Total charge on 48 g ${Mg}^{+2}=\text{no. of}{e}^{-}s\times \text{charge on 1}{e}^{-}=(2.4\times {10}^{24})\times (1.6\times {10}^{-19})=3.84\times {10}^{5}C\approx 3.86\times {10}^{5}C$

Hence, the correct answer is $3.86\times {10}^{5}C$.