4.4 g of CO_2 and 2.24 L of H_2 at text{STP | vedclass

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(B) $1$. Calculate moles of $CO_2$: Molar mass of $CO_2 = 12 + 2 	imes 16 = 44 \ g/mol$. Moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.$2$

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$1.2044 	imes 10^{23}$

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(B) $1$. Calculate moles of $CO_2$: Molar mass of $CO_2 = 12 + 2 	imes 16 = 44 \ g/mol$. Moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.$2$. Calculate moles of $H_2$: At $	ext{STP}$,$1 \ mol$ of gas occupies $22.4 \ L$. Moles of $H_2 = \frac{2.24 \ L}{22.4 \ L/mol} = 0.1 \ mol$.$3$. Total moles of molecules = $0.1 \ mol (CO_2) + 0.1 \ mol (H_2) = 0.2 \ mol$.$4$. Total number of molecules = $	ext{Total moles} 	imes N_A = 0.2 	imes 6.022 	imes 10^{23} = 1.2044 	imes 10^{23}$.

$4.4 \ g$ of $CO_2$ and $2.24 \ L$ of $H_2$ at $\text{STP}$ are mixed in a container. The total number of molecules present in the container will be :

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