Chapter 6, Problem 11A

Interpretation Introduction

Interpretation: The mass of cobalt having same number of atoms as 57.0 g of fluorine needs to be determined.

Concept introduction: The molar mass is defined as mass of a substance in 1 mol. The number of units of the substance in 1 mol is $6.02\times {10}^{23}$ . This is Avogadro’s number and denoted by symbol $N_A$ .

Answer to Problem 11A

  $\text{176}\text{.8 g}$

Explanation of Solution

The mass of fluorine is 57.0 g. The molar mass is 18.99 g/mol thus, number of moles of F will be:

  $n=\frac{m}{M}=\frac{57\text{ g}}{18.99\text{ g/mol}}=3\text{ mol}$

The number of atoms in 1 mol of F will be $6.02\times {10}^{23}$ . Thus, the number of atoms of F in 3 mol will be:

  $\begin{array}{l}N=n\times 6.02\times {10}^{23}\\ =3\times 6.02\times {10}^{23}\\ =1.806\times {10}^{24}\end{array}$

Thus, the number of atoms of fluorine is $1.806\times {10}^{24}$ .

Now, consider the number of atoms of cobalt be $1.806\times {10}^{24}$ thus, number of moles of cobalt will be 3 mol.

The molar mass of cobalt is 58.93 g/mol. Thus, mass of cobalt can be calculated as follows:

  $m=n\times M=3\text{ mol}\times 58.93\text{ g/mol=176}\text{.8 g}$ .

Conclusion

Thus, the mass of cobalt is $\text{176}\text{.8 g}$ .