Question

5.85 g $NaCl$ is dissolved in 1 I. water. The number of ions of $N{a}^{+}$ and $C{I}^{-}$ in 1 mL of this solution will be:

• A. $6.02\times {10}^{19}$
• B. $1.2\times {10}^{22}$
• C. $1.2\times {10}^{20}$
• D. $6.02\times {10}^{20}$

Answer 1

The correct option is C $1.2\times {10}^{20}$

Molarity =$\frac{5.85}{58.5\times 1}$= 0.1M

No. of moles of ions $N{a}^{+}$ , $C{l}^{-}$ = molarity $\times$ volume

$=0.1\times {10}^{-3}={10}^{-4}$

no. of $N{a}^{+}+C{l}^{-}$ ions $={10}^{-4}\times 6.02\times {10}^{23}$

$=6.02\times {10}^{19}$

total ions $=$ no. of $N{a}^{+}$ ions $+$ no. of $C{l}^{-}$ ions

$=2\times 6.02\times {10}^{19}$

$=1.2\times {10}^{20}$