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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    To observe an elevation of boiling point of \[0.05{{\,}^{o}}C,\] the amount of a solute (mol. wt. = 100) to be added to 100 g of water \[({{K}_{b}}=0.5)\] is

    A)  2 g    

    B)  0.5 g   

    C)  1 g     

    D)  0.75 g

    Correct Answer: C

    Solution :

     Elevation of boiling point, \[\Delta {{\Tau }_{b}}=\frac{w\times {{K}_{b}}\times 1000}{M\times W}\] (Here, w and W = weights of solute and solvent respectively, M = molecular weight of solute and \[{{K}_{b}}=\]constant) On substituting values, we get \[0.05=\frac{w\times 0.5\times 1000}{100\times 100}\] or \[w=\frac{0.05\times 100\times 100}{0.5\times 1000}=1g\]


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