248 g of ethylene glycol (C_2H_6O_2) is adde | vedclass

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(C) Step $1$: Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$.$M = (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 24 + 6 + 32 = 62 \ g \ mol

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$20$

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(C) Step $1$: Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$.$M = (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 24 + 6 + 32 = 62 \ g \ mol^{-1}$.Step $2$: Calculate the number of moles of ethylene glycol.$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{248 \ g}{62 \ g \ mol^{-1}} = 4 \ mol$.Step $3$: Calculate the molality $(m)$ of the solution.$m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{4 \ mol}{0.2 \ kg} = 20 \ m$.

$248 \ g$ of ethylene glycol $(C_2H_6O_2)$ is added to $200 \ g$ of water to prepare antifreeze. What is the molality of the resultant solution (in $m$)? $(C=12 \ u, H=1 \ u, O=16 \ u)$

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