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Answer 1 · 2016-10-02T13:12:46

Given

$V \to Volume of the gaseous solute = 4 L$ $V->"Volume of the gaseous solute"=4L$

$P \to Pressure of the gaseous solute = 748 m m = \frac{748}{760} a t m$ $P->"Pressure of the gaseous solute"=748mm=748/760atm$

$T \to Pressure of the gaseous solute = 27^{\circ} C = 300 K$ $T->"Pressure of the gaseous solute"=27^@C=300K$

$n \to Number of moles of the gaseous solute = ?$ $n->"Number of moles of the gaseous solute"=?$

$R \to Universal gas constant = 0.082 L a t m K^{- 1} m o l^{- 1}$ $R->"Universal gas constant"=0.082LatmK^-1mol^-1$

By equation of state for ideal gas

$P V = n R T$ $PV=nRT$

$\Rightarrow n = \frac{P V}{R T} = \frac{\frac{748}{760} \times 4}{0.082 \times 300} = 0.16 m o l$ $=>n=(PV)/(RT)=(748/760xx4)/(0.082xx300)=0.16mol$

Mass of solvent (benzene) $= 58 g = 0.058 k g$ $=58g=0.058kg$

Molality of the solution

$= \frac{moles of solute}{mass of solvent in kg}$ $="moles of solute"/"mass of solvent in kg"$

$= \frac{0.16}{0.058} = 2.76 m$ $=0.16/0.058=2.76m$

The depression of freezing point

$DeltaT_f="molality"xxK_f$

$=2.76xx5.12=14.13^@$