The freezing point of a solution containing 4.8g of a compound in 60g of benzene is 4.48∘C. What is the molar mass of the compound?
Given Kf for benzene =5.1Km−1
and freezing point of pure benzene is 5.5∘C
A
100gmol−1
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B
200gmol−1
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C
300gmol−1
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D
400gmol−1
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Solution
The correct option is D400gmol−1 weight of solute=4.8g weight of solvent=60g Freezing point of solution =4.48∘C Freezing point of solvent =5.5∘C Kf=5.1Km−1
Change in Temp = Freezing point of solvent - Freezing pt.of soln (ΔTf)=(T∘f)−(Tf)=(5.5∘C−4.48∘C)=1.02∘C ΔTf=Kf×m m=1.025.1=0.2 m=No. of moles of non- volatile soluteMass of solvent (in kg) 0.2=4.8×100060×M⇒M=4.8×10000.2×60=400gmol−1