Liquids A and B form an ideal solution in the entire composition range . At 350 K , the vapor pressures of pure A and pure B are7\times 10^{3}Pa\: \: and\: \: 12\times 10^{3}Pa , respectively . The composition of the vapor in equilibrium with a solution containing 40 mol percent of A at this temperature is :

  • Option 1)

    x_{A}=0.37; \: x_{B}=0.63

  • Option 2)

    x_{A}=0.28; \: x_{B}=0.72

  • Option 3)

    x_{A}=0.4; \: x_{B}=0.6

  • Option 4)

    x_{A}=0.76; \: x_{B}=0.24

Answers (1)
A admin

 

Mole Fraction -

\dpi{100} Mole\: Fraction= \frac{Moles\: o\! f\: solute}{Moles\: o\! f solute+Moles\: o\! f solvent}

-

 

 

Vapour Pressure -

It is defined as pressure exerted by vapours on liquid surface at equilibrium and condensation.

-

 

 

 

Negative deviation from Raoult's Law behaviour -

When the pressure exerted by vapours of mixture is less than that in case of ideal behaviour.

- wherein

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

 

 

 

Factors affecting vapour pressure -

V.P.\: \alpha \: T\! emp

- wherein

It depends only on temperature and nature of liquid.

as we have learned in partial pressure

we know that

y_{a} = \frac{P_{A}}{P_{total}} = \frac{P_{A}^{0}\times A}{P_{A}^{0}\times A\times P_{B}^{0}\times B}

y_{A} = \frac{7\times 10^{3}\times 0.4}{7\times 10^{3}\times 0.4 + 12\times 10^{3}\times 0.6}= \frac{2.8}{10} = 0.28

y_{B} = 0.72

 

 


Option 1)

x_{A}=0.37; \: x_{B}=0.63

Option 2)

x_{A}=0.28; \: x_{B}=0.72

Option 3)

x_{A}=0.4; \: x_{B}=0.6

Option 4)

x_{A}=0.76; \: x_{B}=0.24

Exams
Articles
Questions