K_f for water is 1.86 , K , kg , mol^{-1}. If | vedclass

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$.Given values: $\Delta T_f = 2.8 \, K$,$K_f = 1.86 \, K \, kg \,

Vedclass · Accepted Answer

$93$

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$.Given values: $\Delta T_f = 2.8 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$,$i = 1$ (since ethylene glycol is a non-electrolyte).Mass of solvent (water) $= 1.0 \, kg$.Let the mass of solute (ethylene glycol) be $x \, g$.Molar mass of ethylene glycol $(C_2H_6O_2)$ $= (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \, g \, mol^{-1}$.Molality $(m)$ $= \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{x / 62}{1} = \frac{x}{62} \, mol \, kg^{-1}$.Substituting into the equation: $2.8 = 1 	imes 1.86 	imes \frac{x}{62}$.Solving for $x$: $x = \frac{2.8 	imes 62}{1.86} = 93.33 \, g \approx 93 \, g$.

$K_f$ for water is $1.86 \, K \, kg \, mol^{-1}$. If your automobile radiator holds $1.0 \, kg$ of water,how many $gm$ of ethylene glycol $(C_2H_6O_2)$ must you add to lower the freezing point of the solution to $-2.8 \, ^oC$?

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