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NEET Chemistry Solutions / विलयन Question Bank Self Evaluation Test - Solutions

  • question_answer
    Benzene freezes at \[5.50{}^\circ C\]. If the freezing point of 2.48 g of phosphorous in 100 g benzene is \[4.48{}^\circ C\], the atomicity of phosphorus in benzene is (\[{{K}_{f}}\](benzene)\[=5.12\,K\,kgmo{{l}^{-1}}\]):

    A) 1                     

    B) 3     

    C) 4                                 

    D) 8

    Correct Answer: C

    Solution :

    [c] Molecular mass of phosphorous \[=\frac{1000\times {{K}_{f}}\times {{W}_{solute}}}{\Delta {{T}_{f}}\times {{W}_{benzene}}}\] \[=\frac{1000\times 5.12\times 2.48}{1.02\times 100}=124.5={{(P)}_{x}}\] Or , \[x=\frac{124.5}{31}=4\]


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