Question

The number of $C{l}^{-}$ ions in $100mL$ of $0.001MHCl$ 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 $C{l}^{-}$ ions.
$\therefore 0.001M$ 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 $1L$
or in $1000mL$, the number of molecules present
$=6.022\times 10^{20}$
$\therefore$ in $100mL$, the number of molecules present
$=\frac{6.022\times 10^{20}}{1000}\times 100$
$=6.022\times 10^{19}$ molecules