Question

Calculate the ratio of atoms present in 5g of magnesium and 5g of iron.[atomic mass of Mg=24u,Fe=56u]

Answer 1

Answer:

7:3

Explanation:

No of atoms=No. of moles × atomicity × N(6.02×10^23)

No. of Mg atoms = (5/24)×1×N

No. of Fe atoms= (5/56)×1×N

ratio(Mg:Fe)= (5/24)×1×N/ (5/56)×1×N

=7/3

hence answer is 7:3

Answer 2

The ratio of atoms present in 5g of magnesium and 5g of iron is $\frac{7}{3}$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

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

a) $\text{Number of moles of magnesium}=\frac{5g}{24g/mol}$

1 mole of magnesium contains = $6.023\times 10^{23}$ atoms

Thus $\frac{5g}{24g/mol}$ moles of magnesium contains = $\frac{6.023\times 10^{23}}{1}\times \frac{5g}{24g/mol}$ atoms

b) $\text{Number of moles of iron}=\frac{5g}{56g/mol}$

1 mole of iron contains = $6.023\times 10^{23}$ atoms

Thus $\frac{5g}{56g/mol}$ moles of iron contains = $\frac{6.023\times 10^{23}}{1}\times \frac{5g}{56g/mol}$ atoms

Ratio of of atoms present in 5g of magnesium and 5g of iron =$\frac{\frac{5}{24}}{\frac{5}{56}}=\frac{7}{3}$

Learn More about avogadro's law

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