Question

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

Answer 1

The molar mass of $\mathrm{NaCl}$ = The atomic weight of Sodium$\left(\mathrm{Na}\right)$ + The atomic weight of Chlorine$\left(\mathrm{Cl}\right)$

• The atomic weight of $\mathrm{Na}$ = $22.99\mathrm{g}$
• The atomic weight of $\mathrm{Cl}$= $35.45\mathrm{g}$

Thus, the molar mass of $\mathrm{NaCl}$ = $23+35.5=58.5\mathrm{g}/\mathrm{mole}$

Number of moles = $\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}$

Number of moles = $\frac{5.85}{58.5}$

Number of moles = $0.1\mathrm{mole}$

Each $\mathrm{NaCl}$ particle contains one ${\mathrm{Na}}^{+}$ and one ${\mathrm{Cl}}^{-}$ ion, a total of 2 ions

Total moles of ions = $0.1\times 2=0.2\mathrm{moles}$

Number of ions = $\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{ions}\times \mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{no}$

Number of ions = $0.2\times 6.022\times {10}^{23}$

The number of ions present in 5.85 g of sodium chloride = $1.2042\times {10}^{23}\mathrm{ions}$.