2. Solved Papers
  3. VMMC

VMMC VMMC Medical Solved Paper-2006

  • question_answer
    The molar freezing point constant for water is \[1.86{{\,}^{o}}\text{C/mol}\text{.}\] 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)  \[-1.86{{\,}^{o}}C\]                       

    B)         \[1.86{{\,}^{o}}C\]

    C)         \[-3.92{{\,}^{o}}C\]                       

    D)         \[2.42{{\,}^{o}}C\]

    Correct Answer: A

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner