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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Lowering of vapour pressure

  • question_answer
    The vapour pressure lowering caused by the addition of 100 g  of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at \[{{25}^{o}}C\] is 23.8 mm Hg [RPET 1999]

    A)                 1.25 mm Hg         

    B)                 0.125 mm Hg

    C)                 1.15 mm Hg         

    D)                 00.12 mm Hg

    Correct Answer: B

    Solution :

             Given molecular mass of sucrose = 342            Moles of sucrose \[=\frac{100}{342}=0.292\] mole            Moles of water \[N=\frac{1000}{18}=55.5\] moles and            Vapour pressure of pure water \[{{P}^{0}}=23.8\] mm Hg            According to Raoult?s law \[\frac{\Delta P}{{{P}^{0}}}=\frac{n}{n+N}\Rightarrow \frac{\Delta P}{23.8}=\frac{0.292}{0.292+55.5}\]                 \[\Delta P=\frac{23.8\times 0.292}{55.792}=0.125\]mm Hg.


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