The mass of sucrose to be added to 300 g of water to lower its vapour pressure by 1.0 mmHg at 25°C is (Given: p∗( water )=23.8mmHg

The mass of sucrose to be added to 300 g of water to lower its vapour pressure by 1.0 mmHg at 25°C is (Given: p( water )=23.8mmHg

  1. A

    249.9 g

  2. B

    329.4 g

  3. C

    215.2 g

  4. D

    342.2 g

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    Solution:

    Since. p=x1p1 we get

    x1=pp1=23.81.023.8=0.958 Now x1=n1n1+n2=(300/18)mol(300/18)mol+m/342gmol1

    Equating these two, we get

    (300/18)(300/18)+(m/342g)=0.958 or m={(300/18)(10.958)}(342g)0.958=249.89g

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