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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Depression of freezing point of the solvent

  • question_answer
    The molar freezing point constant for water is \[{{1.86}^{o}}C\,mol{{e}^{-1}}\]. If 342 gm of canesugar  \[({{C}_{12}}{{H}_{22}}{{O}_{11}})\] are dissolved in 1000 \[gm\] of water, the solution will freeze at [NCERT 1977; CPMT 1989; Roorkee 2000; DCE 2004]

    A)                 \[-{{1.86}^{o}}C\]            

    B)                 \[{{1.86}^{o}}C\]

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

    D)                 \[{{2.42}^{o}}C\]

    Correct Answer: A

    Solution :

                \[\Delta {{T}_{f}}=1.86\times \left( \frac{342}{342} \right)={{1.86}^{o}}\];  \ \[{{T}_{f}}=-{{1.86}^{o}}C\].


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