Question

Compute the number of ions present in $5.85\text{g}$ of sodium chloride.
$(\text{Given},N_A=6.022\times {10}^{23},\text{atomic mass of}Na=23\text{u},\text{atomic mass of}Cl=35.5\text{u})$

• A. $1.20\times {10}^{23}\text{ions}$
• B. $2.20\times {10}^{14}\text{ions}$
• C. $1.20\times {10}^{30}\text{ions}$
• D. $3.20\times {10}^{16}\text{ions}$

Answer 1

The correct option is A $1.20\times {10}^{23}\text{ions}$

We know that,
$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$
$\text{Number of moles of}NaCl=\frac{\text{5.85}}{\text{58.5}}=0.1\text{mol}$
Each $NaCl$ particle contains $2$ ions; $N{a}^{+}$ and $C{l}^{-}$.
Total moles of ions = $0.1\times 2=0.2\text{mol}$
$\text{One mole}=6.022\times {10}^{23}\text{ions}$
Total number of ions = $0.2\times 6.022\times {10}^{23}=1.2024\times {10}^{23}\text{ions}$