A chloro compound A:(i) forms aldehydes on oz | vedclass

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(B) At $STP$,$22400 \ mL$ of gas corresponds to $1 \ mole$.Given $448 \ mL$ of $A$ weighs $1.53 \ g$.Therefore,the molar mass of $A = \frac{1.53 \ g}{

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$3$

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(B) At $STP$,$22400 \ mL$ of gas corresponds to $1 \ mole$.Given $448 \ mL$ of $A$ weighs $1.53 \ g$.Therefore,the molar mass of $A = \frac{1.53 \ g}{448 \ mL} 	imes 22400 \ mL/mol = 76.5 \ g/mol$.Let the formula be $C_n H_m Cl$.The molar mass is $12n + m + 35.5 = 76.5$,so $12n + m = 41$.Since ozonolysis of $A$ yields aldehydes,$A$ must be an alkene (e.g.,$CH_3-CH=CH-Cl$ or similar structure).For $n=3$,$12(3) + m = 41 \Rightarrow m = 5$.The compound is $C_3H_5Cl$ $(M.W. = 36 + 5 + 35.5 = 76.5)$.Thus,the number of carbon atoms is $3$.

$A$ chloro compound $A$:
$(i)$ forms aldehydes on ozonolysis followed by hydrolysis.
$(ii)$ when vaporized completely,$1.53 \ g$ of $A$ gives $448 \ mL$ of vapour at $STP$.
The number of carbon atoms in a molecule of compound $A$ is ...... .

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$A$ chloro compound $A$:$(i)$ forms aldehydes on ozonolysis followed by hydrolysis.$(ii)$ when vaporized completely,$1.53 \ g$ of $A$ gives $448 \ mL$ of vapour at $STP$.The number of carbon atoms in a molecule of compound $A$ is ...... .