JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in \[\text{mol}\,\text{k}{{\text{g}}^{-1}}\]) of the aqueous solution is                                                            [JEE Main Held on 12-4-2019 Morning]

    A) \[13.88\times {{10}^{1}}\]  

    B) \[13.88\times {{10}^{2}}\]

    C) \[13.88\]          

    D) \[13.88\times {{10}^{3}}\]

    Correct Answer: C

    Solution :

    \[{{X}_{solvent}}=0.8\] If\[{{n}_{T}}=1\] \[{{n}_{Solvent}}=0.8\] \[{{n}_{Solute}}=0.2\] \[molality=\frac{0.2}{\frac{0.8\times 18}{1000}}=13.88\] 


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