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AMU Medical AMU Solved Paper-2006

  • question_answer
    The molar freezing point constant for water is\[-{{1.86}^{o}}C\]. 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                 \[=\frac{342}{342}=1\,m\] We know that,  \[\Delta {{T}_{f}}={{K}_{f\,}}.\,\,m\]                                 \[=1.86\times 1={{1.86}^{o}}\] Hence, freezing point of solution                 \[=0.00-(1.86)=-{{186}^{o}}C\]


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