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NEET Chemistry Solutions / विलयन Question Bank Self Evaluation Test - Solutions

  • question_answer
    Freezing point of an aqueous solution is \[-0.186{}^\circ C.\] If the values of \[{{K}_{b}}\] and \[{{K}_{f}}\] of water are respectively \[0.52K\text{ }kg\text{ }mo{{l}^{-1}}\] and \[1.86K\text{ }kg\text{ }mo{{l}^{-1}}\], then the elevation of boiling point of the solution in K is

    A) 0.52                 

    B) 1.04  

    C) 1.34                 

    D) 0.052

    Correct Answer: D

    Solution :

    [d] \[\Delta {{T}_{f}}=i.{{K}_{f}}.m;\Delta {{T}_{b}}=i.{{K}_{b}}.m\] \[\frac{\Delta {{T}_{f}}}{\Delta {{T}_{b}}}=\frac{{{K}_{f}}}{{{K}_{b}}}\] \[\Delta {{T}_{f}}=0-\left( -0.186{}^\circ C \right)=0.186{}^\circ C\] \[\frac{0.186}{\Delta {{T}_{b}}}=\frac{1.86}{0.52}\Rightarrow \Delta {{T}_{b}}=\frac{0.52\times 0.186}{1.86}=0.052\]


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