Pure water freezes at 273 K and 1 bar. The | vedclass

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(C) The molality $(m)$ of the solution is calculated as:$m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{34.5 \ g / 46 \ g \ m

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$A, B$

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(C) The molality $(m)$ of the solution is calculated as:$m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{34.5 \ g / 46 \ g \ mol^{-1}}{0.5 \ kg} = \frac{0.75 \ mol}{0.5 \ kg} = 1.5 \ m$.The depression in freezing point $(\Delta T_f)$ is given by:$\Delta T_f = K_f 	imes m = 2 \ K \ kg \ mol^{-1} 	imes 1.5 \ mol \ kg^{-1} = 3 \ K$.The new freezing point of the solution is $273 \ K - 3 \ K = 270 \ K$.In the given plots,the freezing point is the temperature where the vapour pressure curve of the solution intersects the vapour pressure curve of ice. For a freezing point of $270 \ K$,the intersection must occur at $270 \ K$. Plot $(A)$ shows the intersection at $270 \ K$,and Plot $(C)$ also shows the intersection at $270 \ K$.

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