From 44 mg of CO_2,2 times 10^{20} molecules | vedclass

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(C) Step $1$: Calculate the total number of molecules in $44 \ mg$ of $CO_2$. Molar mass of $CO_2 = 12 + 2 	imes 16 = 44 \ g/mol$. Mass of $CO_2 = 44

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$4 	imes 10^{20}$

Vedclass · Answer

(C) Step $1$: Calculate the total number of molecules in $44 \ mg$ of $CO_2$. Molar mass of $CO_2 = 12 + 2 	imes 16 = 44 \ g/mol$. Mass of $CO_2 = 44 \ mg = 44 	imes 10^{-3} \ g$. Number of moles $= \frac{44 	imes 10^{-3} \ g}{44 \ g/mol} = 10^{-3} \ mol$. Total molecules $= 10^{-3} 	imes 6 	imes 10^{23} = 6 	imes 10^{20}$ molecules. Step $2$: Subtract the removed molecules from the total. Molecules left $= (6 	imes 10^{20}) - (2 	imes 10^{20}) = 4 	imes 10^{20}$ molecules.

From $44 \ mg$ of $CO_2$,$2 \times 10^{20}$ molecules of $CO_2$ are removed. The number of molecules of $CO_2$ left are (Given $N_A = 6 \times 10^{23}$):-

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