If the density of a 2 M solution of ethylene | vedclass

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(C) Density of solution $= 1.11 \ g \ mL^{-1}$.Molarity of solution $= 2 \ M$, which means $2 \ 	ext{moles}$ of solute are present in $1000 \ mL$ of

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$2.02$

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(C) Density of solution $= 1.11 \ g \ mL^{-1}$.Molarity of solution $= 2 \ M$, which means $2 \ 	ext{moles}$ of solute are present in $1000 \ mL$ of solution.Mass of solution $= 	ext{density} 	imes 	ext{volume} = 1.11 \ g/mL 	imes 1000 \ mL = 1110 \ g$.Molar mass of ethylene glycol $(C_2H_6O_2) = 62 \ g/mol$.Mass of solute $= 	ext{moles} 	imes 	ext{molar mass} = 2 \ mol 	imes 62 \ g/mol = 124 \ g$.Mass of solvent $= 	ext{Mass of solution} - 	ext{Mass of solute} = 1110 \ g - 124 \ g = 986 \ g$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg} = \frac{2 \ mol}{0.986 \ kg} \approx 2.02 \ m$.

If the density of a $2 \ M$ solution of ethylene glycol in water is $1.11 \ g/mL$, the molality (in $m$) of the solution is approximately

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