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  • How to solve this problem- A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg forpentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for
pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be

  • Option 1)

    0.549

  • Option 2)

    0.200

  • Option 3)

    0.786

  • Option 4)

    0.478

 

Answers (1)

best_answer

As we learned

 

Raoult's Law -

The partial pressure of any volatile constituents of a solution at a given temperature is equal to the product of vapour pressure of pure constituent and its mole fraction in the solution.

P_{A}=P_{A}^{0}x_{A}

P_{B}=P_{B}^{0}x_{B}

 

 

- wherein

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

P_{A}  and P_{B} are vapour pressures of A and B respectively.

 

 P_{T}=P_{p}^{0}x_{p}+P_{h}^{0}x_{h}= 440\times \frac{1}{5}+120\times \frac{4}{5}

=88+96=184; P_{p}^{0}x_{p}=y_{p}P_{T}; \frac{88}{184}=y_{p}

y_{p}=0.478


Option 1)

0.549

Option 2)

0.200

Option 3)

0.786

Option 4)

0.478

Posted by

satyajeet

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