Question

Calculate the number of atoms present in $71g$ of ${Cl}_2$.

• A. $1.205\times {10}^{-24}$ atoms
• B. $1.205\times {10}^{23}$ atoms
• C. $1.205\times {10}^{24}$ atoms
• D. $1.205\times {10}^{-23}$ atoms

Answer 1

The correct option is C $1.205\times {10}^{24}$ atoms

Weight of $C{l}_2=71$ g

Molecular weight of $C{l}_2$ $=71$ g/mol

Number of moles of $C{l}_2$ $=\frac{71g}{71g/gatom}=1$ moles.

The Avogadro's number is $6.023\times {10}^{23}$ molecules/mol.

The number of atoms of $C{l}_2$ $=2$ atoms $\times 1$ mol $\times 6.023\times {10}^{23}$ molecules/mol
$=1.205\times {10}^{24}$ atoms