Maximum number of atoms are present in

11.2 litres of SO2 at STP

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22.4 litres of Helium at STP

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2.0 g. of hydrogen

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11.2 litres of methane at STP

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Solution

The correct option is D
11.2 litres of methane at STP

No. of atoms in $11.2 l i t .$ of $S O_{2}$ = $\frac{11.2}{22.4} \times N_{A} \times 3 = 1.5 N$

No. of Atoms in $2 g$ $H_{2}$ = $\frac{2}{2} \times N_{A} \times 2 = 2 N$

No. of Atoms in $22.4 l i t .$ of $H e$ = $\frac{22.4}{22.4} \times N_{A} = N$

No. of Atoms in $11.2 l i t$ of $C H_{4}$ = $\frac{11.2}{22.4} \times N_{A} \times 5$ = $2.5 N$

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Similar questions

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Maximum number of atoms are present in

The correct option is D 11.2 litres of methane at STP No. of atoms in 11.2 lit. of SO2 = 11.222.4 × NA × 3 = 1.5N No. of Atoms in 2g H2 = 22 × NA × 2 = 2N No. of Atoms in 22.4 lit. of He = 22.422.4 × NA = N No. of Atoms in 11.2 lit of CH4 = 11.222.4 × NA × 5 = 2.5 N