An antifreeze solution is prepared from 222.6 | vedclass

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(A) The molar mass of ethylene glycol $[C_2H_4(OH)_2]$ is $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g/mol$. Molality $(m)$ is defined as th

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$17.95$

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(A) The molar mass of ethylene glycol $[C_2H_4(OH)_2]$ is $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g/mol$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Moles of ethylene glycol $= \frac{222.6 \ g}{62 \ g/mol} = 3.59 \ mol$. Mass of solvent (water) $= 200 \ g = 0.2 \ kg$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{3.59 \ mol}{0.2 \ kg} = 17.95 \ m$.

An antifreeze solution is prepared from $222.6 \ g$ of ethylene glycol $[C_2H_4(OH)_2]$ and $200 \ g$ of water. What is the molality of this solution (in $m$)?

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