  Question

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

A
2.4×1023C  B
6.82×105C  C
3.86×105C  D
1.93×105C  Solution

The correct option is C $$3.86\times { 10 }^{ 5 }C$$Molar mass of $${Mg}^{+2} = 24 g$$Given mass of $${Mg}^{+2} = 48 g$$No. of moles of $${Mg}^{+2} = \cfrac{\text{Given mass}}{\text{Molar mass}} = \cfrac{48}{24} = 2 \text{ moles}$$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 $${Mg}^{+2} = \text{no. of }{e}^{-}s \times \text{charge on 1 } {e}^{-} = \left( 2.4 \times {10}^{24} \right) \times \left( 1.6 \times {10}^{-19} \right) = 3.84 \times {10}^{5} C \approx 3.86 \times {10}^{5} C$$Hence, the correct answer is $$3.86 \times {10}^{5} C$$.Chemistry

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