A flask contains 98 mg of $\mathrm{H}_2 \mathrm{SO}_4$. If $3.01 \times 10^{20}$ molecules of $\mathrm{H}_2…

A flask contains 98 mg of $\mathrm{H}_2 \mathrm{SO}_4$. If $3.01 \times 10^{20}$ molecules of $\mathrm{H}_2 \mathrm{SO}_4$ are removed from the flask, the number of moles of $\mathrm{H}_2 \mathrm{SO}_4$ remained in flask is $\left(\mathrm{N}=6.02 \times 10^{23}\right)$
  1. $1 \times 10^{-4}$
  2. $5 \times 10^{-4}$
  3. $1.66 \times 10^{-4}$
  4. $9.95 \times 10^{-3}$

Solution

Weight of $\mathrm{H}_2 \mathrm{SO}_4(W)=98 \mathrm{mg}$
$=98 \times 10^{-3} \mathrm{~g}\left[1 \mathrm{mg}=10^{-3} \mathrm{~g}\right]$
No. of mole in flask $\left(\mathrm{n}_1\right)=\frac{98 \times 10^{-3}}{98}=10^{-3}$ mole
No. of mole removed from the flask $\left(n_2\right)=\frac{3.01 \times 10^{20}}{6.023 \times 10^{23}}$ $=\frac{1}{2} \times 10^{-3}$ $\therefore$ Remaining no. of moles in the flask $\left(\mathrm{n}_1-\mathrm{n}_2\right)$ $\begin{aligned} & =1 \times 10^{-3}-\frac{1}{2} \times 10^{-3} \\ & =0.5 \times 10^{-3} \\ & =5 \times 10^{-4} \mathrm{~mole} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

Practice more Some Basic Concepts of Chemistry questions on Aicharya