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NEET AIPMT SOLVED PAPER 1999

  • question_answer
                    If 0.15 g of solute, dissolved in 15 g of solvent, is boiled at a temperature higher by \[{{0.216}^{\text{o}}}C,\] than that of the pure solvent, the molecular weight of the substance is (molal elevation constant for the solvent is \[{{2.16}^{\text{o}}}C\]):

    A)                                                                                                                                                                                              1.01          

    B)                         10             

    C)                         10.1          

    D)              100

    Correct Answer: D

    Solution :

                    w = 0.15 g,   W = 15 g,   \[\Delta {{T}_{b}}={{0.216}^{o}}C,\]                 \[{{K}_{b}}=2.16,\,\,m=?\]                 \[m=\frac{1000\times {{K}_{b}}}{\Delta {{T}_{b}}}=\frac{w}{W}\] \[=\frac{1000\times 2.16\times 0.15}{0.216\times 15}=100\]


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