If Avogadro's number is $A_{0}$ , the number of sulphur atoms present in $200 m L$ of $1 N$ $H_{2} {S O}_{4}$ is

A_{0} / 5

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A_{0} / 2

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A_{0} / 10

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A_{0}

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Solution

The correct option is D $A_{0} / 10$
Normality $= n_{t} \times$ Molarity

$1 = 2 \times M$

$M = \frac{1}{2} m o l L^{- 1} = 0.5 M H_{2} S O_{4}$

Moles of $H_{2} S O_{4} = 0.5 \times 0.2 = 0.1 m o l e s$

1 mole $H_{2} S O_{4}$ contains 1 mole of sulphur.

Hence, number of moles of the sulphur atom in 0.1 moles $H_{2} S O_{4}$ = 0.1 mole of sulphur.

The number of sulphur atoms on 0.1 mole $= 0.1 \times A_{0} \times 1$

$= \frac{A_{0}}{10}$

Therefore, option C is correct.

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