Question

The number of electrons present in $100ml$ of $0.1N$ $H_2{SO}_4$ is:

• A. $6.85\times {10}^{22}$
• B. $7.50\times {10}^{23}$
• C. $1.5\times {10}^{23}$
• D. $1.8\times {10}^{22}$

Answer 1

The correct option is C $1.5\times {10}^{23}$

Normality=$n\times Molarity$

$\therefore 0.1=2\times \frac{molesof{H}_{2}S{O}_4}{0.1}$

$\therefore Molesof{H}_{2}S{O}_4=0.005moles$

$1$ molecule of $H_{2}S{O}_4$ contains $50$ electrons.

No. of molecules in $0.005$ moles=$0.005\times 6.022\times {10}^{23}$

=$0.03011\times {10}^{23}$ molecules

$\therefore$ No. of electrons in $0.005$ moles=$0.03011\times {10}^{23}\times 50$

=$1.5\times {10}^{23}$ electrons