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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Critical Thinking Solutions

  • question_answer
    A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at \[{{25}^{o}}C\]. Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at \[{{25}^{o}}C\]. Calculate the molecular weight of the solute                                             [UPSEAT 2001]

    A)                 74.2       

    B)                 75.6

    C)             67.83     

    D)                 78.7

    Correct Answer: C

    Solution :

             We have,                    \[\frac{{{p}^{0}}-21.85}{21.85}=\frac{30\times 18}{90\times m}\], for I case         .....(i)                    wt. of solvent \[=90+18=108gm\]                    \[\frac{{{p}^{0}}-22.15}{22.15}=\frac{30\times 18}{108\times m}\], for II case       .....(ii)                    By eq. (1) \[p_{m}^{0}-21.85m=21.85\times 6=131.1\]                    By eq. (2) \[p_{m}^{0}-22.15m=22.15\times 5=110.75\]                    0.30m = 20.35                                 \[m=\frac{20.35}{0.30}=67.83\]


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