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An aqueous solution of 2% non-volatile solute exerts pressure of 1.004 bar at the normal boiling point of solvent. The molar mass of the solute is g/mol is    

  • Option 1)

    85.42

  • Option 2)

    41.35

  • Option 3)

    90

  • Option 4)

    94.42

 

Answers (1)

best_answer

As learnt in

Rault's Law -

The total vapour pressure of binary mixture of miscible liquids be having ideally is given by

P_{T}= P_{A}^{0}x_{A}+P_{B}^{0}x_{B}

Where x_{A}  and  x_{B} are mole fraction of A and B in liquid phase.

 

 

- wherein

P_{A}^{0}  and P_{B}^{0} are vapour pressures of pure liquids.

 

 \frac{p{_{1}}^{0}-p_{1}}{p{_{1}}^{0}}= \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}

\frac{1.013\cdot 1.004}{1.013}= \frac{2\times 18}{M_{2}\times 98}

\Rightarrow M_{2} = \frac{2\times 18\times 0.013}{0.009\times 98} = 41.35 g\ mol^{-1}


Option 1)

85.42

This option is incorrect

Option 2)

41.35

This option is correct

Option 3)

90

This option is incorrect

Option 4)

94.42

This option is incorrect

Posted by

prateek

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