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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    The degree of dissociation of XY in case of following reaction is found to be \[0.4\] from vapour density measurement. If the observed vapour density of X Y be 100 then its molecular weight is \[XY\to X+Y\]

    A)  \[140\]            

    B)  \[70\]

    C)  \[280\]

    D)  \[210\]

    Correct Answer: C

    Solution :

                    The degree of dissociation \[(\alpha )=\frac{{{D}_{T}}-{{D}_{0}}}{{{D}_{0}}}\] \[0.4=\frac{{{D}_{T}}-100}{100}\] \[{{D}_{T}}=40+100=140\]                 \[\therefore \]  \[{{D}_{T}}=\frac{mol.\,wt.}{2}\]                 \[\because \]     Mol. wt. \[={{D}_{T}}\times 2=140\times 2\]                                                 \[=280\]


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