Maximum number of atoms are present in Options: A . &nbsp 11.2 l i t r e s of S O 2 at STP B . &nbsp 22.4 l i t r e s of Helium at STP C . &nbsp 2.0 g . of hydrogen D . &nbsp 11.2 l i t r e s of methane at STP

Answer: Option D : D No. of atoms in 11.2 l i t . of S O 2 = 11.2 22.4 × N A × 3 = 1.5 N No. of Atoms in 2 g H 2 = 2 2 × N A × 2 = 2 N No. of Atoms in 22.4 l i t . of H e = 22.4 22.4 × N A = N No. of Atoms in 11.2 l i t of C H 4 = 11.2 22.4 × N A × 5 = 2.5 N

Maximum number of atoms are present in

Discussion Forum : Mole Concept

Question -

Maximum number of atoms are present in

Options:

A .

11.2 l i t r e s

S O_{2}

at STP

B .

22.4 l i t r e s

of Helium at STP

C .

2.0 g

. of hydrogen

D .

11.2 l i t r e s

of methane at STP

Answer: Option D
:
D

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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