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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    The vapour pressure of water at \[{{20}^{o}}C\] is 17.5 mm Hg. If 18 g of glucose \[({{C}_{6}}{{H}_{12}}{{O}_{6}})\] is added to 178.2 g of water at \[{{20}^{o}}C\], the vapour pressure of the resulting solution will be       AIEEE  Solved  Paper-2007

    A) 16.500 mm Hg

    B)        17.325 mm Hg

    C)        17.675 mm Hg

    D)        15.750 mm Hg

    Correct Answer: B

    Solution :

                    Moles of glucose \[=\frac{18}{180}=0.1\], moles of \[{{H}_{2}}O=\frac{1782}{18}=9.9\] \[\frac{{{P}^{o}}-{{P}_{S}}}{{{P}_{S}}}=\frac{moles\text{ }of\text{ }glucose}{moles\text{ }of\text{ }water}\,\, & ;\,\,\,17.5-{{P}_{S}}\]\[=\frac{0.1\times {{P}_{S}}}{9.9}\] \[{{P}_{S}}=17.325\] mm Hg.


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