$22$ grams of ${C O}_{2}$ at STP is identical to :

1

mole of

{C O}_{2}

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6.022 \times 10^{23}

atoms

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6.022 \times 10^{23}

molecules

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11.2

liters

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22.4

liters

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Solution

The correct option is D $11.2$ liters
The molecular weigh of $C O_{2}$ is 44 g/mole.
Twenty-two grams of $C O_{2}$ at STP corresponds to $\frac{22}{44} = 0.5$ moles.
1 mole of $C O_{2} = 6.022 \times 10^{23}$ molecules.
0.5 mole of $C O_{2} = 3.011 \times 10^{23}$ molecules.
1 mole of $C O_{2} =$ 3 moles of atoms whic corresponds to $3 \times 6.022 \times 10^{23}$ . atoms.
0.51 mole of $C O_{2} =$ 1.5 moles of atoms whic corresponds to $1.5 \times 6.022 \times 10^{23}$ . atoms.
At STP, 1 mole of any gas occupies a volume of 22.4 L.
Hence, at STP, 0.5 mole of $C O_{2}$ gas occupies a volume of $0.5 \times 22.4 = 11.2$ L.

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Similar questions

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6.022 \times 10^{23}

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22 grams of CO2 at STP is identical to :

$22$ grams of ${C O}_{2}$ at STP is identical to :