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)

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

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