If $3.01 \times 10^{20}$ molecules are removed from $98 \mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$,…

If $3.01 \times 10^{20}$ molecules are removed from $98 \mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$, then the number of moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ left are
  1. $0.1 \times 10^{-3}$
  2. $0.5 \times 10^{-3}$
  3. $1.66 \times 10^{-3}$
  4. $9.95 \times 10^{-2}$

Solution

Moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $98 \mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$
$=\frac{1}{98} \times 0.098=0.001$
Moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ removes $=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}}=0.5 \times 10^{-3}=0.0005$
Moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ left $=0.001-0.0005=0.5 \times 10^{-3}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more SOME BASIC CONCEPTS OF CHEMISTRY questions on Aicharya