If Avogadro's number is A $_{0}$ , the number of sulphur atoms present in 200 mL of 1N H $_{2}$ SO $_{4}$ 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 C
A $_{0}$ /10

Avogadro's number = A $_{0}$
Normality = n $_{t}$ $\times$ Molarity
1= 2 $\times$ M
M= $\frac{1}{2} m o l L^{- 1}$
Moles of H $_{2}$ SO $_{4}$ = $\frac{1}{2}$ $\times$ 0.2 = 0.1 moles 21
Normality = 1; Volume = 200 ml (0.2 litre)
Moles of hydrogen = 0.2 moles
Moles of sulphur = 0.1 moles
Atoms = 0.1A $_{0}$

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A_{0}

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$s i n n x = \sum_{r = 0}^{n} a_{r} s i n^{r} x$ where n is an odd natural number, then

\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} - \dots + {(- 1)}^{n} \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}}

\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}

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