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(C) $1$. Calculate the freezing point of the solution:$\Delta T_{f} = K_{f} 	imes m = 2.0 \; K \; kg \; mol^{-1} 	imes 0.5 \; mol \; kg^{-1} = 1.0 \

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$2.18$

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(C) $1$. Calculate the freezing point of the solution:$\Delta T_{f} = K_{f} 	imes m = 2.0 \; K \; kg \; mol^{-1} 	imes 0.5 \; mol \; kg^{-1} = 1.0 \; K$.Freezing point of solution $T = 273 - 1 = 272 \; K$.$2$. Calculate the initial pressure of the gas:Using $PV = nRT$,$P = \frac{nRT}{V} = \frac{0.1 \; mol 	imes 0.08 \; dm^{3} \; atm \; K^{-1} \; mol^{-1} 	imes 272 \; K}{1.0 \; dm^{3}} = 2.176 \; atm$.$3$. After withdrawing the stoppers,the piston is frictionless and exposed to atmospheric pressure $(P_{ext} = 1 \; atm)$. The gas will expand until its internal pressure equals the external pressure.Using Boyle's Law $(P_{1}V_{1} = P_{2}V_{2})$:$2.176 \; atm 	imes 1.0 \; L = 1.0 \; atm 	imes V_{2}$.$V_{2} = 2.176 \; L \approx 2.18 \; L$.

$A$. Ebullioscopic constant	$I$. Depression of freezing point
$B$. Cryoscopic constant	$II$. Total pressure is the sum of partial pressures of the components
$C$. Henry's law	$III$. Elevation of boiling point
$D$. Dalton's law	$IV$. Solubility of a gas in liquid

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