JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Lowering of vapour pressure

  • question_answer
    The vapour pressure of benzene at a certain temperature is \[640\,mm\] of \[Hg\]. A non-volatile and non-electrolyte solid weighing \[2.175g\] is added to \[39.08g\] of benzene. The vapour pressure of the solution is \[600mm\] of \[Hg\]. What is the molecular weight of solid substance [CBSE PMT 1999; AFMC 1999]

    A)                 49.50     

    B)                 59.6

    C)                 69.5       

    D)                 79.8

    Correct Answer: C

    Solution :

             \[\frac{{{P}^{0}}-{{P}_{s}}}{{{P}^{0}}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}\]  \[\because \,\frac{W}{M}>\frac{w}{m}\]  \[\Rightarrow \frac{640-600}{640}\]               \[=\frac{w}{m}\times \frac{M}{W}\Rightarrow \frac{40}{640}=\frac{2.175\times 78}{m\times 39.08}\] ;           \[m=\frac{2.175\times 78}{39.08}\times \frac{640}{40}\]                 \[m=69.45\].


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