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BHU PMT BHU PMT Solved Paper-2002

  • question_answer
    By dissolving 10 g of a non-volatile solute in 100 g of benzene, the boiling point rises by\[1{}^\circ C\]. The molecular mass of solute is : [\[{{k}_{b}}\]for benzene\[=2.53\text{ }k{{m}^{-1}}\]]

    A)  235 g                                    

    B)  352 g

    C)  250 g                                    

    D)  253 g

    Correct Answer: D

    Solution :

                     Key Idea: The molecular mass of solute is calculated by using following formula \[M=\frac{1000\times {{K}_{b}}\times W}{W\times \Delta {{T}_{b}}}\] where M = molecular mass of solute = ? \[{{K}_{b}}=2.53k{{m}^{-1}}\] w = weight of solute = 10 g W = weight of solvent = 100 g \[\Delta {{T}_{b}}=\]elevation in boiling point\[={{1}^{o}}C\] Substituting the values                 \[M=\frac{1000\times 2.53\times 10}{100\times 1}\] \[=253g\]


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