Question

$5.85 \,g$ of $NaCl$ is dissolved in $1 \,L$ of pure water. The number of ions in $1 \,mL$ of this solution is

• 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

Answer: (C)

Number of molecules of $NaCl$ in $1 L$ water

$=\frac{5.85}{58.5} \times 6.023 \times 10^{23}=6.023 \times 10^{22} $ molecules

Number of molecules of $NaCl$ in one $mL$ solution

$=\frac{6.023 \times 10^{22}}{1000}=6.023 \times 10^{19} $

$NaCl \to Na ^{+}+ Cl ^{-}$

One molecule gives $2$ ions.

$\therefore $ Number of ions in one $mL$ solution

$=6.023 \times 10^{19} \times 2=1.2 \times 10^{20}$