JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    18 g glucose \[({{C}_{6}}{{H}_{12}}{{O}_{6}})\]is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is : [JEE Main Solved Paper-2016 ]

    A) 759.0

    B) 7.6

    C) 76.0                                       

    D) 752.4

    Correct Answer: D

    Solution :

                    Assuming temperature to be \[{{100}^{o}}C\] Relative lowering of vapour pressure Equation\[\frac{{{P}^{o}}-{{P}^{5}}}{{{P}^{o}}}={{X}_{solute}}=\frac{n}{n+N}\] Modified forms of equation is\[\frac{{{P}^{o}}-{{P}_{s}}}{{{P}_{s}}}=\frac{n}{N}\] \[{{P}^{o}}=760\]torr \[{{P}_{S}}=?\] \[\frac{760-{{P}_{S}}}{{{P}_{S}}}=\frac{18/180}{178.2/18}\] \[{{P}_{S}}=752.4\]torr


You need to login to perform this action.
You will be redirected in 3 sec spinner