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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The amount of solute (molar mass\[60\text{ }g\text{ }mo{{l}^{-1}}\]) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10%, is

    A)  30 g                                      

    B)  60 g

    C)  120 g                    

    D)         12 g

    E)  24 g

    Correct Answer: B

    Solution :

    Relative lowering of vapour pressure is given by the formula \[\frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{{{w}_{A}}}{{{M}_{A}}}\times \frac{{{M}_{B}}}{{{w}_{B}}}\] As vapour pressure of water is lowered by 10%. \[\therefore \]  \[\frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{10}{100}\] \[\therefore \]  \[\frac{10}{100}=\frac{{{w}_{A}}}{60}\times \frac{18}{180}\] or          \[{{w}_{A}}=60\text{ }g\]


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