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Q.

The number of molecules present in 4.4 g of ${CO}_{2}$ gas is

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a

$6.023 \times 10^{23}$

b

$6.023 \times 10^{22}$

c

$5.023 \times 10^{23}$

d

$6.023 \times 10^{24}$

answer is D.

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Detailed Solution

Number of molecules = Number of moles $\times$ Avogadro number

$= \frac{4.4}{44} \times 6.023 \times 10^{23} = 6.023 \times 10^{22}$

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