Question

The total number of electrons in $18ml$ of water (density $=1gm{l}^{-1}$) is

• A. $6.02\times 10^{23}$
• B. $6.02\times 10^{25}$
• C. $6.02\times 10^{24}$
• D. $6.02\times 10^{23}$

Answer 1

Answer: (C)

In $18mL$, number of moles of

$H_{2}O=\frac{\text{ mass of }{H}_{2}O}{\text{ molecular mass }}$

$=\frac{\text{ density }\times \text{ velume }}{\text{ molecular mass }}$

$=\frac{1\times 18}{18}=1mol$

$\because$ Number of moles of $H_{2}O$ in $1mol$

$=6.022\times 10^{23}$

and number of $e^{-}$ in $1$ molecule of $H_{2}O$

$=1\times 2+8=10$

$\therefore$ Number of $e^{-}$ in $1mole$ of $H_{2}O$

$=6.022\times 10^{23}\times 10$

$=6.022\times 10^{24}$