Question

If $3.01\times {10}^{20}$ molecules are removed from $98\text{mg}$ of $H_{2}S{O}_4$, then the number of moles of $H_{2}S{O}_4$ left will be :

• A. $0.1\times {10}^{-3}$
• B. $1.66\times {10}^{-3}$
• C. $9.95\times {10}^{-2}$
• D. $0.5\times {10}^{-3}$

Answer 1

The correct option is D $0.5\times {10}^{-3}$

Moles of $H_{2}S{O}_4$ in $98\text{mg}\left(0.098\text{g}\right)$ of $H_{2}S{O}_4$
$=\frac{1}{98}\times 0.098=0.001\text{mol}$
Moles of $H_{2}S{O}_4$ removed $=\frac{3.01\times {10}^{20}}{6.02\times {10}^{23}}$
$=0.0005\text{mol}$
Moles of $H_{2}S{O}_4$ left $=0.001-0.0005$
$=0.0005$$=0.5\times {10}^{-3}$