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NEET Chemistry NEET PYQ-Solutions

  • question_answer
    If 0.15 g of solute, dissolved in 15 g of solvent, is boiled at a temperature higher by \[0.216{}^\circ C,\] than that of the pure solvent, the molecular weight of the substance is (molal elevation constant for the solvent is \[2.16{}^\circ C\]):                    [AIPMT 1999]

    A) 1.01

    B) 10

    C) 10.1

    D) 100

    Correct Answer: D

    Solution :

    [d] 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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