Question

The number of $ Cl^{-} $ ions in $ 100\, mL $ of $ 0.001\, M \,HCl $ solution is

• A. $ 6.022 \times 10^{23} $
• B. $ 6.022 \times 10^{20} $
• C. $ 6.022 \times 10^{19} $
• D. $ 6.022 \times 10^{24} $

Answer 1

Answer: (C)

$1M$ of $HCl$ solution contains
$6.022 \times 10^{23}$ molecules of $Cl^-$ ions.
$\therefore 0.001 \,M$ solution will contain
$= 6.022 \times 10^{23} \times 10^{-3}$ molecule
$= 6.002 \times 10^{20}$ molecule
So, $6.022 \times 10^{20}$ molecule are present in $1 \,L$
or in $1000 \,mL$, the number of molecules present
$= 6.022 \times 10^{20}$
$\therefore $ in $100 \,mL$, the number of molecules present
$ = \frac{6.022\times 10^{20}}{1000} \times 100$
$= 6.022 \times 10^{19}$ molecules