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VMMC VMMC Medical Solved Paper-2004

  • question_answer
    If 0.15 g of a solute, dissolved is 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 substances (molal elevation constant for the solvent is \[2.16{}^\circ C\]) is:

    A)  100     

    B)                                         10.1                      

    C)  10                         

    D)         1.001

    Correct Answer: A

    Solution :

    \[\omega =0.15\,g,\,W=15\,g,\] \[\Delta \,{{T}_{b}}=0.216\,{{\,}^{o}}C\] \[{{K}_{b}}=2.16{{\,}^{o}}C\]                 \[\therefore \]  \[M=\frac{{{K}_{b}}\times \omega \times 1000}{\Delta {{\Tau }_{b}}\times W}\]                                 \[=\frac{2.16\times 0.15\times 1000}{0.216\times 15}=100\]


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