For CCl4:
Kf=29.8mC∘ , Tf=−22.9C∘
Kb=5.03mC∘ , Tb0=76.8C∘
1) Find molality of a solution
20.0g(C10H8)×128.17g(C10H8)1mole(C10H8)=0.156mol(C10H8)
mmolarity=kg(CCl4)mol(C10H8)=0.1607kg(CCl4)0.156mol(C10H8)=0.917molal
2) Find Tfrsolution
ΔTfr=Kf×m=29.8×0.917=28.94C∘
ΔTfr=Tsolvent−Tsolution
Tsolution=Tsolvent−ΔTfr=−22.9−28.94=−51.84C∘
3) Find Tb(solution)
ΔTb.p.=Kb×m=5.03×0.971=4.88C∘
Tsolution=Tsolvent+ΔTb.p.=76.8+4.88=81.68C∘
4) Find osmotic pressure (at 25C∘)
π=iMRT
i=1
M=molarity=volumesolutionmoles(C10H8)
V=dm=1.25320.0+160.7=144.2ml
M=molarity=144.2×10−30.156moles=1.082M
π=1×1.082×0.08206×298=26.5atm
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