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

The Correct Option is 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 $H _{2} O$ $=6.022 \times 10^{23} \times 10$ $=6.022 \times 10^{24}$