Complete combustion of X g of an organic comp | vedclass

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(A) Mass of $CO_2 = 0.25$ g. Moles of $CO_2 = \frac{0.25}{44} \approx 0.00568$ mol. Mass of $C = 	ext{Moles of } CO_2 	imes 12 = 0.00568 	imes 12 =

Vedclass · Accepted Answer

$273$

Vedclass · Answer

(A) Mass of $CO_2 = 0.25$ g. Moles of $CO_2 = \frac{0.25}{44} \approx 0.00568$ mol. Mass of $C = 	ext{Moles of } CO_2 	imes 12 = 0.00568 	imes 12 = 0.06818$ g. Given percentage of $C = 25\%$. So,$0.06818 = 0.25 	imes X \implies X = \frac{0.06818}{0.25} = 0.2727$ g. Similarly for $H$,Mass of $H_2O = 0.12$ g. Moles of $H_2O = \frac{0.12}{18} \approx 0.00666$ mol. Mass of $H = 0.00666 	imes 2 = 0.01333$ g. Given percentage of $H = 4.89\%$. So,$0.01333 = 0.0489 	imes X \implies X = \frac{0.01333}{0.0489} = 0.2726$ g. Thus,$X \approx 0.273$ g. Expressing as $273 	imes 10^{-3}$ g.

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