JEE Main & Advanced Chemistry Solutions / विलयन Sample Paper Topic Test - Solutions

  • question_answer
    A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g /mol is

    A) 136.2

    B) 171.2   

    C) 68.4

    D) 34.2

    Correct Answer: C

    Solution :

    [c] Two solutions are isotonic if their osmotic pressure are equal
    \[{{\pi }_{1}}={{\pi }_{2}}\]
    \[{{M}_{1}}S{{T}_{1}}={{M}_{2}}S{{T}_{2}}\] (\[{{M}_{1}}\] and \[{{M}_{2}}\] are molarities)
    At a given temperature, \[{{M}_{1}}={{M}_{2}}\]
    \[\underset{\text{Conesugar}}{\mathop{\frac{1000\,{{w}_{1}}}{{{m}_{1}}{{v}_{1}}}}}\,=\underset{\text{Unknown}}{\mathop{\frac{1000\,{{w}_{2}}}{{{m}_{2}}{{v}_{2}}}}}\,\]            \[({{V}_{1}}={{V}_{2}}\,100\,mL)\]
                \[\therefore \]    \[\frac{{{w}_{1}}}{{{m}_{1}}}=\frac{{{w}_{2}}}{{{m}_{2}}}\]
                \[\frac{5}{342}=\frac{1}{{{m}_{2}}}\]
    \[{{m}_{2}}=\frac{342}{5}=68.4\,g\,mo{{l}^{-1}}\]


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