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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    For an aqueous solution, freezing point is\[-0.186{}^\circ C\]. Elevation of the boiling point of the same solution is (\[{{K}_{f}}=1.86{}^\circ mo{{l}^{-1}}kg\]and\[{{K}_{b}}=0.512{}^\circ mo{{l}^{-1}}kg\])

    A)  \[0.186{}^\circ C\]

    B)  \[0.0512{}^\circ C\]

    C)  \[1.86{}^\circ C\]

    D)  \[5.12{}^\circ C\]

    Correct Answer: B

    Solution :

     \[\Delta {{T}_{b}}=m{{K}_{b}}\] \[\Delta {{T}_{f}}=m{{K}_{f}}\] \[\frac{\Delta {{T}_{b}}}{\Delta {{T}_{f}}}=\frac{{{K}_{b}}}{{{K}_{f}}}=\frac{0.512}{1.86}\] \[\Delta {{T}_{b}}=\frac{0.512}{1.86}\times 0.186\] \[=0.0512{}^\circ C\]


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