If the density of a 2.05 M CH_3COOH solution | vedclass

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(C) Molarity $(M)$ = $2.05 \ mol/L$. This means $2.05 \ moles$ of $CH_3COOH$ are present in $1000 \ mL$ of solution.Molar mass of $CH_3COOH$ = $60 \ g

Vedclass · Accepted Answer

$2.28$

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(C) Molarity $(M)$ = $2.05 \ mol/L$. This means $2.05 \ moles$ of $CH_3COOH$ are present in $1000 \ mL$ of solution.Molar mass of $CH_3COOH$ = $60 \ g/mol$.Mass of solute $(CH_3COOH)$ = $2.05 \ mol 	imes 60 \ g/mol = 123 \ g$.Mass of solution = Density $	imes$ Volume = $1.02 \ g/mL 	imes 1000 \ mL = 1020 \ g$.Mass of solvent (water) = Mass of solution - Mass of solute = $1020 \ g - 123 \ g = 897 \ g = 0.897 \ kg$.Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{2.05 \ mol}{0.897 \ kg} \approx 2.28 \ m$.

If the density of a $2.05 \ M$ $CH_3COOH$ solution is $1.02 \ g/mL$,what is the molality of the solution?

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