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Manipal Medical Manipal Medical Solved Paper-2005

  • question_answer
    The molar freezing point constant for water is\[1.86{}^\circ C/mole\]. If 342 g of cane sugar\[({{C}_{12}}{{H}_{22}}{{O}_{11}})\]is dissolved in 1000 g of water, the solution will freeze at:

    A)  \[-1.86{}^\circ C\]     

    B)  \[1.86{}^\circ C\]

    C)  \[-3.92{}^\circ C\]      

    D)  \[2.42{}^\circ C\]

    Correct Answer: A

    Solution :

     Molality of cane sugar solution \[=\frac{342}{342\times 1}=1\,m\] We know that\[\Delta {{T}_{f}}={{K}_{f}}.m\] \[=1.86\times 1\] \[=1.86{}^\circ \] Hence, freezing point of solution \[=0.00-(1.86)=-1.86{}^\circ C\]


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