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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 $(P_{1}V_{1} = P_{2}V_{2})$:$2.176 \; atm 	imes 1.0 \; L = 1.0 \; atm 	imes V_{2}$.$V_{2} = 2.176 \; L \approx 2.18 \; L$.

List-$I$	List-$II$
$(A)$ Azeotrope	$(I)$ $\Delta T_b = i K_b m$
$(B)$ Henry's law	$(II)$ $p = K_H x$
$(C)$ Cryoscopic constant	$(III)$ $\Delta T_f / m$
$(D)$ Van't Hoff factor	$(IV)$ Deviation from Raoult's law
	$(V)$ $\pi = CRT$

Similar Questions

$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{moles}$ of liquid $B$ make a solution having an observed vapour pressure of $42 \, \text{torr}$. The vapour pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively. The described solution:

The osmotic pressure of a $0.5 \ M$ aqueous solution of $CH_3COOH$ having a $pH$ of $2$ at temperature $T$ is . . . . . . . (in $RT$)

Match the following.

List-$I$ List-$II$

$(A)$ Azeotrope $(I)$ $\Delta T_b = i K_b m$

$(B)$ Henry's law $(II)$ $p = K_H x$

$(C)$ Cryoscopic constant $(III)$ $\Delta T_f / m$

$(D)$ Van't Hoff factor $(IV)$ Deviation from Raoult's law

$(V)$ $\pi = CRT$

The correct answer is

After adding a solute,the freezing point of the solution decreases to $-0.186 \ ^{\circ}C$. Calculate $\Delta T_b$ if $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.521 \ K \ kg \ mol^{-1}$. (Assume the freezing point of pure solvent is $0 \ ^{\circ}C$)

$P_A = (235y - 125xy) \, \text{mm of Hg}$. $P_A$ is the partial pressure of $A$,$x$ is the mole fraction of $B$ in the liquid phase in the mixture of two liquids $A$ and $B$,and $y$ is the mole fraction of $A$ in the vapour phase. Then $P^o_B$ in $\text{mm of Hg}$ is:

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