Grignard reagent RMgBr (P) reacts with water | vedclass

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Question

(A) At $STP$,$22.4 \ dm^3$ of a gas corresponds to $1 \ mole$. Given that $1.4 \ dm^3$ of gas $(Q)$ weighs $1 \ g$. Therefore,the molar mass of $(Q)$

Vedclass · Accepted Answer

$6$

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(A) At $STP$,$22.4 \ dm^3$ of a gas corresponds to $1 \ mole$. Given that $1.4 \ dm^3$ of gas $(Q)$ weighs $1 \ g$. Therefore,the molar mass of $(Q)$ is $\frac{22.4 \ dm^3/mol 	imes 1 \ g}{1.4 \ dm^3} = 16 \ g/mol$. Since the gas $(Q)$ is formed from $RMgBr$ and water,it is an alkane. The alkane with molar mass $16 \ g/mol$ is methane $(CH_4)$. Thus,the alkyl group $R$ in $RMgBr$ is a methyl group $(CH_3)$,and $(P)$ is $CH_3MgBr$. Reaction of $CH_3MgBr$ with dry ice $(CO_2)$ followed by acidic hydrolysis $(H_3O^{+})$ yields acetic acid $(CH_3COOH)$ as compound $(Z)$. The molar mass of $CH_3COOH$ is $12 + 3(1) + 12 + 2(16) + 1 = 60 \ g/mol$. Weight of $0.1 \ mole$ of $(Z)$ = $0.1 \ mol 	imes 60 \ g/mol = 6 \ g$.

Grignard reagent $RMgBr$ $(P)$ reacts with water and forms a gas $(Q)$. One gram of $Q$ occupies $1.4 \ dm^3$ at $STP$. $(P)$ on reaction with dry ice in dry ether followed by $H_3O^{+}$ forms a compound $(Z)$. $0.1 \ mole$ of $(Z)$ will weigh . . . . . . $g$. (Nearest integer)

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