On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressure of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol-1 and of octane = 114 g mol-1 )

  • Option 1)

    144.5 kP

  • Option 2)

    72.0 kP

  • Option 3)

    36.1 kP

  • Option 4)

    96.2 kPa

 

Answers (1)

As we 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.

 

 p=p_{A}^{0}\chi_{A}+p_{B}^{0}\chi_{B}

p_{A}^{0}=\frac{25/100}{25/100+35/114}=0.4488

p_{B}^{0}=1-\chi_{A}=0.5512

p=105\times 0.4488+45\times 0.5512

p = 71.92

Correct option is 2.

 


Option 1)

144.5 kP

This is an incorrect option.

Option 2)

72.0 kP

This is the correct option.

Option 3)

36.1 kP

This is an incorrect option.

Option 4)

96.2 kPa

This is an incorrect option.

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