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JIPMER Jipmer Medical Solved Paper-2015

  • question_answer
    Vapour pressure of benzene at \[30{}^\circ C\] is 121.8 mm when 15 g of a non-volatile solute is dissolved in 250 g of benzene, its vapour pressure is decreased to 120.2 mm. The molecular weight of the solute is:

    A)  35.67 g                

    B)     356.7 g

    C)     432.8 g                

    D)     502.7 g

    Correct Answer: B

    Solution :

                    As, \[\frac{{{P}^{0}}-P}{{{P}^{0}}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}\] \[{{P}^{0}}\]= vap. pressure of pure component \[P\] = vap. pressure in solution \[w\] = mass of solute, \[m\]= mol. wt. of solute \[W\]= mass of solvent, \[M\]= mol. wt. of solvent \[\Rightarrow \]               \[\frac{121.8-120.2}{121.8}=\frac{\frac{15}{m}}{\frac{250}{78}}\] \[\Rightarrow \]               \[m=356.265\,g\]            


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