$A$ flask contains $98 \ mg$ of $H_2SO_4$. If $3.01 \times 10^{20}$ molecules of $H_2SO_4$ are removed from the flask,the number of moles of $H_2SO_4$ remaining in the flask is $\left(N_A = 6.02 \times 10^{23}\right)$
- A$1 \times 10^{-4}$
- B$5 \times 10^{-4}$
- C$1.66 \times 10^{-4}$
- D$9.95 \times 10^{-3}$
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