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If \( 3.01 \times 10^{20} \) molecules are removed from \( 98 \mathrm{mg} \) of \( \mathrm{H}_{2} \mathrm{SO}_{4} \), then number of moles of
\( \mathrm{H}_{2} \mathrm{SO}_{4} \) left are
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\( \mathrm{H}_{2} \mathrm{SO}_{4} \) left are
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Verified Answer
The correct answer is:
\( 0.5 \times 10^{-3} \mathrm{~mol} \)
1 mole of $\mathrm{H}_{2} \mathrm{SO}_{4}=6.022 \times 10^{23}$ molecules of $\mathrm{H}_{2} \mathrm{SO}_{4}=98 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$
$98 \mathrm{mg}=1 \times 10^{-3}$ moles $=6.022 \times 10^{20}$
molecules of $\mathrm{H}_{2} \mathrm{SO}_{4}$
No. of molecules removed $=3.01 \times 10^{20}$
No. of molecules left $=6.023 \times 10^{20}-3.01 \times 10^{20}$
$=6.023 \times 10^{20}-3.01 \times 10^{20}$
$=3.01 \times 10^{20}$ molecules left
$6.022 \times 10^{23}$ molecules of $\mathrm{H}_{2} \mathrm{SO}_{4}=1$ mole
1 molecule $=\frac{1}{6.022 \times 10^{23}}$ moles
$3.01 \times 10^{20}$ molecules $=\frac{3.01 \times 10^{20}}{6.023 \times 10^{20}}$
$=0.5 \times 10^{-3}$ mol
$98 \mathrm{mg}=1 \times 10^{-3}$ moles $=6.022 \times 10^{20}$
molecules of $\mathrm{H}_{2} \mathrm{SO}_{4}$
No. of molecules removed $=3.01 \times 10^{20}$
No. of molecules left $=6.023 \times 10^{20}-3.01 \times 10^{20}$
$=6.023 \times 10^{20}-3.01 \times 10^{20}$
$=3.01 \times 10^{20}$ molecules left
$6.022 \times 10^{23}$ molecules of $\mathrm{H}_{2} \mathrm{SO}_{4}=1$ mole
1 molecule $=\frac{1}{6.022 \times 10^{23}}$ moles
$3.01 \times 10^{20}$ molecules $=\frac{3.01 \times 10^{20}}{6.023 \times 10^{20}}$
$=0.5 \times 10^{-3}$ mol
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