4.4 g of $C O_{2}$ and 2.24 litre of $H_{2}$ at STP are mixed in a container. The total number of molecules present in the container will be:

6.022 \times 10^{23}

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1.2044 \times 10^{23}

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12.044 \times 10^{23}

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6.023 \times 10^{24}

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Solution

The correct option is C $1.2044 \times 10^{23}$
$4.4 g$ of $C O_{2} = 0.1$ mole

1 mole occupies 22.4 L
$∴$ 2.24 L equals $0.1$ mole

Total number of moles $= 0.1 + 0.1 = 0.2$ mole
1 mole occupies $N_{A}$ molecules

$∴$ The total number of molecules present in the container will be = $0.2 \times N_{A} =$ $1.2044 \times 10^{23}$ .

Hence, the correct option is $B$

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4.4 g of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of molecules present in the container will be:

The correct option is C 1.2044×10234.4g of CO2=0.1 mole1 mole occupies 22.4 L∴ 2.24 L equals 0.1 moleTotal number of moles =0.1+0.1=0.2 mole1 mole occupies NA molecules∴ The total number of molecules present in the container will be = 0.2×NA= 1.2044×1023.Hence, the correct option is B

4.4 g of $C O_{2}$ and 2.24 litre of $H_{2}$ at STP are mixed in a container. The total number of molecules present in the container will be: