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What volume of $2 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ is required to form $0.2 \mathrm{~N}$ of $100 \mathrm{~mL}$ of solution?
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The correct answer is:
$5 \mathrm{~mL}$
$\because \mathrm{M}-\mathrm{H}_{2} \mathrm{SO}_{4}=2 \mathrm{~N}-\mathrm{H}_{2} \mathrm{SO}_{4}$
$\therefore 2 \mathrm{M}-\mathrm{H}_{2} \mathrm{SO}_{4}=4 \mathrm{~N}-\mathrm{H}_{2} \mathrm{SO}_{4}$
$N_{1} V_{1}=N_{2} V_{2}$
$4 \times V_{1}=0.2 \times 100$
$V_{1}=\frac{0.2 \times 100}{4}$
$=5 \mathrm{~mL}$
$\therefore 2 \mathrm{M}-\mathrm{H}_{2} \mathrm{SO}_{4}=4 \mathrm{~N}-\mathrm{H}_{2} \mathrm{SO}_{4}$
$N_{1} V_{1}=N_{2} V_{2}$
$4 \times V_{1}=0.2 \times 100$
$V_{1}=\frac{0.2 \times 100}{4}$
$=5 \mathrm{~mL}$
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