Molar mass of the hydrocarbon (X) which on oz | vedclass

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(B) The ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.Ethanal is $CH_3CHO$ and propanone is $(CH_3

Vedclass · Accepted Answer

$70$

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(B) The ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.Ethanal is $CH_3CHO$ and propanone is $(CH_3)_2CO$.By joining these two fragments at the carbonyl carbons,we get the structure of the alkene $(X)$:$CH_3-CH=C(CH_3)_2$.The chemical formula of this hydrocarbon is $C_5H_{10}$.The molar mass of $C_5H_{10} = (5 	imes 12) + (10 	imes 1) = 60 + 10 = 70 \ g\ mol^{-1}$.

Molar mass of the hydrocarbon $(X)$ which on ozonolysis consumes one mole of $O_3$ per mole of $(X)$ and gives one mole each of ethanal and propanone is $...... g\ mol^{-1}$ (Molar mass of $C: 12\ g\ mol^{-1}, H: 1\ g\ mol^{-1}$)

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