Question

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

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

Answer 1

Answer: (C)

In $18\, mL$, number of moles of

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

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

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

$\because$ Number of moles of $H _{2} O$ in $1\, mol$

$=6.022 \times 10^{23}$

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

$=1 \times 2+8=10$

$\therefore $ Number of $e^{-}$ in $1\, mole$ of $H _{2} O$

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

$=6.022 \times 10^{24}$