A sample of 4.5 g of an unknown monohydric a | vedclass

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(A) The reaction between a monohydric alcohol and methylmagnesium iodide is: $ROH + CH_3MgI \rightarrow ROMgI + CH_4(g)$.According to the stoichiometr

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(A) The reaction between a monohydric alcohol and methylmagnesium iodide is: $ROH + CH_3MgI ightarrow ROMgI + CH_4(g)$.According to the stoichiometry,$1 \ mole$ of $ROH$ produces $1 \ mole$ of $CH_4$ gas.At $NTP$ conditions,$1 \ mole$ of gas occupies $22.4 \ L$.Given volume of $CH_4 = 3.1 \ L$.Moles of $CH_4 = \frac{3.1}{22.4} \approx 0.1384 \ mol$.Since moles of $ROH = 	ext{moles of } CH_4$,we have $0.1384 = \frac{	ext{mass}}{	ext{Molar Mass}} = \frac{4.5}{M.M}$.$M.M = \frac{4.5}{0.1384} \approx 32.51 \ g/mol$.The nearest integer is $33$.

$A$ sample of $4.5 \ g$ of an unknown monohydric alcohol,$ROH$,was added to methylmagnesium iodide. $A$ gas is evolved,collected,and its volume is measured to be $3.1 \ L$ at $NTP$. The molecular weight of the unknown alcohol is $g/mol$. [Nearest integer]

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