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If Avogadro's number is $\mathrm{A}_{0}$, the number of sulphur atoms present in $200 \mathrm{~mL}$ of $1 \mathrm{~N}$ $\mathrm{H}_{2} \mathrm{SO}_{4}$ is
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The correct answer is:
$\mathrm{A}_{0} / 10$
Avogadro's number $=\mathrm{A}_{0}$
Normality $=\mathrm{n}_{\mathrm{t}} \times$ Molarity
$1=2 \times \mathrm{M}$
$\mathrm{M}=\frac{1}{2} \mathrm{~mol} \mathrm{~L}^{-1}$
Moles of $\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{1}{2} \times 0.2=0.1 \mathrm{moles}$
Normality $=1 ;$ Volume $=200 \mathrm{ml}(0.2 \mathrm{litre})$
Moles of hydrogen $=0.2$ moles
Moles of sulphur $=0.1$ moles
Atoms $=0.1 \mathrm{~A}_{0}$
Normality $=\mathrm{n}_{\mathrm{t}} \times$ Molarity
$1=2 \times \mathrm{M}$
$\mathrm{M}=\frac{1}{2} \mathrm{~mol} \mathrm{~L}^{-1}$
Moles of $\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{1}{2} \times 0.2=0.1 \mathrm{moles}$
Normality $=1 ;$ Volume $=200 \mathrm{ml}(0.2 \mathrm{litre})$
Moles of hydrogen $=0.2$ moles
Moles of sulphur $=0.1$ moles
Atoms $=0.1 \mathrm{~A}_{0}$
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