A triatomic gas having $3.6132 \times 10^{24}$ atoms has to be stored in a container.What should be the volume of container to store the gas at $1$ atm pressure?

67.2 L

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33.6 L

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44.8 L

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56 L

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Solution

The correct option is C $44.8 L$
Number of atoms in Triatomic gas $= 3.6132 \times 10^{24} a t o m s$
Number of triatomic molecules in container $= \frac{3.6132 \times 10^{24}}{3} = 1.2044 \times 10^{24}$
Number of moles $= \frac{1.2044 \times 10^{24}}{6.023 \times 10^{23}} = 2 m o l e s$

Volume of 1 mole gas $= 22.4 L$ at STP

Volume of 2 mole gas $= 22.4 \times 2 = 44.8 L$

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A triatomic gas having 3.6132×1024 atoms has to be stored in a container.What should be the volume of container to store the gas at 1 atm pressure?

The correct option is C 44.8LNumber of atoms in Triatomic gas=3.6132×1024 atomsNumber of triatomic molecules in container =3.6132×10243=1.2044×1024Number of moles=1.2044×10246.023×1023=2molesVolume of 1 mole gas =22.4L at STPVolume of 2 mole gas =22.4×2=44.8L

A triatomic gas having $3.6132 \times 10^{24}$ atoms has to be stored in a container.What should be the volume of container to store the gas at $1$ atm pressure?