Answer : The correct option is, (a) \(\frac{24}{184}\times 100\)
Solution : Given,
Molar mass of \(MgBr_2\) = 184 g/mole
Molar mass of Mg = 24 g/mole
Formula used :
\(\%\text{ Composition of Mg}=\frac{\text{Molar mass of Mg}}{\text{Molar mass of }MgBr_2}\times 100\)
Now put all the given values in this formula, we get the percent composition of magnesium.
\(\%\text{ Composition of Mg}=\frac{24g/mole}{184g/mole}\times 100\)
\(\%\text{ Composition of Mg}=\frac{24}{184}\times 100\)
Therefore, the correct answer is, (a)
