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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    12g of a nonvolatile solute dissolved in 108g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the lute is:                

    A)                 80                          

    B)                                        60                          

    C)                 20                                          

    D)                        40

    Correct Answer: C

    Solution :

                    From Raoult's law                 Relative lowering in vapour pressure                 \[\Delta p=\frac{{{p}^{0}}-p}{{{p}^{0}}}=\frac{n}{N}=\frac{w}{m}\times \frac{M}{W}\]                 \[w=12\,g;\,\,W=108\,g,\] \[m=?;\]                 \[M=18\,g,\,\,\Delta p=0.1\]                 \[\Delta p=\frac{w}{m}\times \frac{M}{W}\]                 \[0.1=\frac{12}{m}\times \frac{18}{108}\]                             \[m=\frac{12\times 18}{10.8}=20\]                


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