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NEET AIPMT SOLVED PAPER SCREENING 2006

  • question_answer
                    1.00 g of a non-electrolyte solute (molar mass \[250g\,mo{{l}^{-1}}\]) was dissolved in 51.2 g of benzene. If the freezing point depression constant, \[{{K}_{f}}\] of benzene is \[5.12\,K\,kg\,mo{{l}^{-1}},\] the freezing point of benzene will be lowered by:

    A)                                                                                                                                                                                             0.4 K

    B)                  0.3 K    

    C)                  0.5 K                    

    D)                  0.2 K

    Correct Answer: A

    Solution :

                              Molality of non-electrolyte solute                 \[=\frac{\frac{weight\,of\,solute\,in\,gram}{molecular\,weight\,of\,solute}}{weight\,of\,solvent\,in\,kg}\]                 \[=\frac{\frac{1}{250}}{0.0512}=\frac{1}{250\times 0.0512}=0.0781\,m\]                 \[\Delta {{T}_{f}}={{k}_{f}}\times molality\,of\,solution\]                 \[=5.12\times 0.0781=0.4\,K\]


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