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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    When any 2 g solute dissolves in 50 g water than boiling point comes\[100.5{}^\circ C\]. If\[{{K}_{b}}=0.5\] then molecular weight will be

    A)  4                

    B)  40

    C)  80              

    D)  100

    Correct Answer: B

    Solution :

     \[{{T}_{b}}={{K}_{b}}\times m\] \[={{K}_{b}}\times \frac{w}{M}\times \frac{1000}{W}\] \[M=\frac{{{K}_{b}}\times w\times 1000}{\Delta {{T}_{b}}\times W}\] \[=\frac{0.5\times 2\times 1000}{0.5\times 50}\] \[=40\]


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