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

  • question_answer
    Freezing point of an aqueous solution is\[\left( -0.186 \right){}^\circ C\]. Elevation of boiling point of the same solution is\[{{K}_{b}}=0.512{}^\circ C,K{{  }_{f}}=1.86{}^\circ C\], find the increase in boiling point.

    A) \[0.186{}^\circ C\]                   

    B) \[0.0512{}^\circ C\]

    C) \[0.092{}^\circ C\]                   

    D) \[0.2372{}^\circ C\]

    Correct Answer: B

    Solution :

    [b] \[\Delta {{T}_{b}}={{K}_{b}}\frac{{{W}_{B}}}{{{M}_{B}}\times {{W}_{A}}}\times 1000;\] \[\Delta {{T}_{f}}={{K}_{f}}\frac{{{W}_{B}}}{{{M}_{B}}\times {{W}_{A}}}\times 1000;\] \[\frac{\Delta {{T}_{b}}}{\Delta {{T}_{f}}}=\frac{{{K}_{b}}}{{{K}_{f}}}=\frac{\Delta {{T}_{b}}}{-0.186}=\frac{0.512}{1.86}=0.0512{}^\circ C.\]


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