When 5 g of acetic acid is dissolved in 1 L | vedclass

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(D) Mass of solute $(CH_3COOH)$ = $5 \ g$. Molar mass of $CH_3COOH$ = $60 \ g/mol$. Moles of solute = $\frac{5}{60} \approx 0.0833 \ mol$. Volume of s

Vedclass · Accepted Answer

$0.106$

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(D) Mass of solute $(CH_3COOH)$ = $5 \ g$. Molar mass of $CH_3COOH$ = $60 \ g/mol$. Moles of solute = $\frac{5}{60} \approx 0.0833 \ mol$. Volume of solvent (ethanol) = $1000 \ mL$. Mass of solvent = $	ext{Density} 	imes 	ext{Volume} = 0.789 \ g/mL 	imes 1000 \ mL = 789 \ g = 0.789 \ kg$. Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.0833 \ mol}{0.789 \ kg} \approx 0.1056 \ m \approx 0.106 \ m$.

When $5 \ g$ of acetic acid is dissolved in $1 \ L$ of ethanol,assuming no reaction occurs between them,calculate the molality of the resulting solution. (Density of ethanol = $0.789 \ g/mL$) (in $m$)

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