The density of a 2.05 M solution of acetic a | vedclass

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(B) Given: Molarity $(M) = 2.05 \ mol \ L^{-1}$,Density $(d) = 1.02 \ g \ mL^{-1}$.Molar mass of acetic acid $(CH_3COOH) = 12 + 3 + 12 + 16 + 16 + 1 =

Vedclass · Accepted Answer

$2.28$

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(B) Given: Molarity $(M) = 2.05 \ mol \ L^{-1}$,Density $(d) = 1.02 \ g \ mL^{-1}$.Molar mass of acetic acid $(CH_3COOH) = 12 + 3 + 12 + 16 + 16 + 1 = 60 \ g \ mol^{-1}$.Mass of solute in $1 \ L$ solution = $2.05 \ mol 	imes 60 \ g \ mol^{-1} = 123 \ g$.Mass of $1 \ L$ solution = $Volume 	imes Density = 1000 \ mL 	imes 1.02 \ g \ mL^{-1} = 1020 \ g$.Mass of solvent (water) = $Mass \ of \ solution - Mass \ of \ solute = 1020 \ g - 123 \ g = 897 \ g = 0.897 \ kg$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{2.05 \ mol}{0.897 \ kg} \approx 2.28 \ mol \ kg^{-1}$.

The density of a $2.05 \ M$ solution of acetic acid $(CH_3COOH)$ in water is $1.02 \ g \ mL^{-1}$. The molality of the solution is ......... $mol \ kg^{-1}$.

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