Question

The number of molecules and atoms present in $142g$ of chlorine gas is:

• A. $6\times {10}^{23}$ molecules, $6\times {10}^{23}$ atoms
• B. $1.2\times {10}^{24}$ molecules, $2.4\times {10}^{24}$ atoms
• C. $2.4\times {10}^{24}$ molecules, $1.2\times {10}^{24}$ atoms
• D. $1.2\times {10}^{24}$ molecules, $1.2\times {!0}^{24}$ atoms

Answer 1

Answer: (B)

$1.2\times {10}^{24}$ molecules, $2.4\times {10}^{24}$ atoms

$2\times 35.5g$ of $Cl=71g$ of $C{l}_2$ has $6.02\times {10}^{23}$ molecules.

Hence, $142g$ of $C{l}_2$ gas has $\frac{142}{71}\times 6.022\times {10}^{23}=1.20\times {10}^{24}$ molecules.

Each $C{l}_2$ molecules has two chlorine atoms.

$\therefore 1.2\times {10}^{24}\times 2=2.4\times {10}^{24}$ $atoms.$