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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    An aqueous solution freezes at \[-{{0.186}^{o}}C\] \[({{k}_{f}}={{1.86}^{o}};\,{{k}_{b}}={{0.512}^{o}})\]What is the elevation in boiling point?

    A)  \[0.186\]                            

    B)  \[0.512\]

    C)  \[\frac{0.512}{1.86}\]                   

    D)  \[0.0512\]

    Correct Answer: D

    Solution :

    \[\Delta {{T}_{f}}=k_{f}^{'}\times m\] \[0.186=1.86\times m\] \[m=\frac{0.186}{1.86}=0.1mol\]                 \[\therefore \]  \[\Delta {{T}_{b}}={{k}_{b}}\times molality\]                                 \[\Delta {{T}_{b}}=0.512\times 0.1\]                                 \[\Delta {{T}_{b}}=0.0512\]


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