Calculate the ratio of atoms present in $5 g$ of magnesium and $5 g$ of iron (Atomic Mass $Mg = 24, Fe = 56$ ).

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Solution

We know, Number of moles of $Mg$ = Given Mass/ Molecular mass
= $\frac{5}{24}$
Number of moles of $Fe$ = $\frac{5}{56}$
Now, 1 mole contains = $6.022 \times 10^{23}$ atoms
$\frac{5}{24}$ moles of $Mg$ contains = $\frac{5}{24} \times 6.022 \times 10^{23}$ atoms
$\frac{5}{56}$ moles of $Fe$ contains = $\frac{5}{56} \times 6.022 \times 10^{23}$ atoms
The ratio of atoms present in $5 g Mg$ and $5 g$ $Fe$ = $\frac{\frac{5}{24} \times 6.022 \times 10^{23}}{\frac{5}{56} \times 6.022 \times 10^{23}} = \frac{56}{24}$
Hence, $\frac{R a t i o o f a t o m s p r e s e n t i n 5 g M g}{R a t i o o f a t o m s p r e s e n t i n 5 g F e} = 2.33$

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