Question

Compute the number of ions present in $5.85$ g of sodium chloride.

• A. $1.2042\times {10}^{23}$ ions
• B. $1.2042\times {10}^{26}$ ions
• C. $1.2042\times {10}^{29}$ ions
• D. $1.2042\times {10}^{30}$ ions

Answer 1

The correct option is A $1.2042\times {10}^{23}$ ions

According to given data,
5.85 g of $NaCl=\frac{5.85}{58.5}=0.1$ moles of NaCl particle
Each NaCl particle contains 2 ions $N{a}^{+}$ and $C{l}^{-}$
$\Rightarrow$ Total moles of ions $=0.1\times 2$ $=0.2$ moles
No. of ions $=0.2\times 6.022\times {10}^{23}$ $=1.2042\times {10}^{23}$ ions