A solution containing 62 g of ethylene glyco | vedclass

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(C) The molar mass of ethylene glycol $(C_2H_6O_2)$ is $62 \ g \ mol^{-1}$.Moles of ethylene glycol $= \frac{62 \ g}{62 \ g \ mol^{-1}} = 1 \ mol$.Let

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$64$

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(C) The molar mass of ethylene glycol $(C_2H_6O_2)$ is $62 \ g \ mol^{-1}$.Moles of ethylene glycol $= \frac{62 \ g}{62 \ g \ mol^{-1}} = 1 \ mol$.Let the mass of water remaining in the liquid phase be $w \ g$.The freezing point depression is $\Delta T_f = 0 - (-10) = 10 \ K$.Using the formula $\Delta T_f = K_f 	imes m$,where $m$ is the molality:$10 = 1.86 	imes \frac{1 \ mol}{(w / 1000) \ kg}$.$w = \frac{1.86 	imes 1000}{10} = 186 \ g$.The amount of water separated as ice $= 	ext{Initial mass} - 	ext{Remaining mass} = 250 \ g - 186 \ g = 64 \ g$.

$A$ solution containing $62 \ g$ of ethylene glycol in $250 \ g$ of water is cooled to $-10 \ ^\circ C$. If $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$,the amount of water (in $g$) separated as ice is:

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