When 0.01 mol of an organic compound contain | vedclass

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(C) Let $M$ be the molar mass of the compound in $g \ mol^{-1}$.Mass of $0.01 \ mol$ of the compound $= 0.01 	imes M \ g$.Since the compound contains

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$200$

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(C) Let $M$ be the molar mass of the compound in $g \ mol^{-1}$.Mass of $0.01 \ mol$ of the compound $= 0.01 	imes M \ g$.Since the compound contains $60 \%$ carbon by mass,the mass of carbon in the compound $= 0.01 	imes M 	imes 0.60 = 0.006 \ M \ g$.Moles of carbon in the compound $= \frac{0.006 \ M}{12} = 0.0005 \ M$.During combustion,all carbon in the compound is converted to $CO_2$.Moles of $CO_2$ produced $= \frac{4.4 \ g}{44 \ g \ mol^{-1}} = 0.1 \ mol$.Since $1 \ mol$ of $CO_2$ contains $1 \ mol$ of carbon,the moles of carbon in the compound must equal the moles of $CO_2$ produced.Therefore,$0.0005 \ M = 0.1$.$M = \frac{0.1}{0.0005} = 200 \ g \ mol^{-1}$.

When $0.01 \ mol$ of an organic compound containing $60 \%$ carbon was burnt completely,$4.4 \ g$ of $CO_2$ was produced. The molar mass of the compound is $......... \ g \ mol^{-1}$ (Nearest integer).

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