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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    \[12\,\,g\] of a non-volatile solute dissolved in \[108\,\,g\] of water produces the relative lowering of vapour pressure of\[0.1\]. The molecular mass of the solute is

    A) \[80\]                                   

    B) \[60\]

    C)  \[20\]                                  

    D)  \[40\]

    Correct Answer: C

    Solution :

    From Raoult's law, relative lowering in vapour pressure                 \[\Delta p=\frac{{{p}^{o}}-p}{{{p}^{o}}}=\frac{n}{N}=\frac{W}{m}\times \frac{M}{W}\] Here,     \[w=12\,\,g;\,\,W=108\,\,g,\,\,m=?\]                 \[M=18\,\,g,\,\,\Delta p=0.1\]                 \[\Delta p=\frac{W}{m}\times \frac{M}{W}\] or            \[0.1=\frac{12}{m}\times \frac{18}{108}\]                 \[m=\frac{12\times 18}{10.8}=20\]


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