Question

The number of iron (Fe) atoms present in a pure sample of solid iron with a mass of 10 grams is equal to- (The atomic mass of iron is 55.9) .

• A. $\left(10.0\right)\times \left(55.9\right)\times (6.02\times {10}^{23})$ atoms
• B. $\frac{(6.02\times {10}^{23})}{\left(10.0\right)\left(55.9\right)}$
• C. $\frac{\left(10.0\right)(6.02\times {10}^{23})}{\left(55.9\right)}$
• D. $\frac{\left(55.9\right)}{\left(10.0\right)(6.02\times {10}^{23})}$
• E. $\frac{\left(10.0\right)}{\left(55.9\right)(6.02\times {10}^{23})}$

Answer 1

The correct option is C $\frac{\left(10.0\right)(6.02\times {10}^{23})}{\left(55.9\right)}$

The number of iron (Fe) atoms present in a pure sample of solid iron with a mass of 10 grams is equal to $\frac{\left(10.0\right)(6.02\times {10}^{23})}{\left(55.9\right)}$
Hence, the option (C) is the correct answer.
Note : $\frac{\left(10.0\right)}{\left(55.9\right)}$ represents the ratio of mass of iron to molar mass. It represents number of moles of iron.
When number of moles of iron is multiplied with avogadro's number $(6.02\times {10}^{23})$, we get number of iron atoms. (The atomic mass of iron is $55.9$) .