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  • A non-ideal solution contains 2.5 moles of solute A and 7.5 moles of solvent B. At 298 K, the vapour pressure of pure A is 100 torr and that of pure B is 50 torr. The observed vapour pressure of th

A non-ideal solution contains 2.5 moles of solute A and 7.5 moles of solvent B. At 298 K, the vapour pressure of pure A is 100 torr and that of pure B is 50 torr. The observed vapour pressure of the solution is 130 torr. Calculate the vapour pressure of A and B in the solution assuming negative deviation from Raoult's law.

 

Option: 1

63.63 torr and 66.37 torr


Option: 2

86.67 torr and 43.33 torr
 


Option: 3

150.00 torr and 0 torr

 


Option: 4

30.00 torr and 100.00 torr


Answers (1)

best_answer

According to Raoult's law, the vapour pressure of a component in a solution is given by:

p_A=x_A P_A \text { and } p_B=x_B P_B

where x_A and x_B are the mole fractions of A and B in the solution, and P_A and P_B are the vapour pressures of pure A and B, respectively.

However, in the case of negative deviation, the vapour pressure of the solution is lower than that predicted by Raoult's law. This happens because the intermolecular forces between unlike molecules are stronger than those between like molecules.

Let's assume that the observed vapour pressure of the solution is due to the contribution of both components, A and B:

p_{\text {observed }}=p_A+p_B=x_A P_A+x_B P_B

We can rearrange this equation to solve for x_A:

x_A=\frac{p_{\text {observed }}-x_B P_B}{P_A-P_B}

Substituting the given values, we get:

x_A=\frac{130-0.75 \times 50}{100-50}=0.3

Therefore, x_B=0.7

Now, we can calculate the vapour pressure of A and B in the solution using the following equations:

\begin{aligned} &p_A=x_A P_A e^{-\frac{\Delta H}{R T} x_B}\\ &p_B=x_B P_B^* e^{-\frac{\Delta H_{\ell} \Delta}{R T} x_A} \end{aligned}


where \Delta H_{AB} and \Delta H_{BA} are the enthalpy of mixing of A and B, and R is the gas constant.

Assuming \Delta H_{A B}=\Delta H_{B A}=-5000 \mathrm{~J} / \mathrm{mol} \text { and } \mathrm{T}=298 \mathrm{~K} \text {, } We get:

\begin{aligned} & p_A=0.3 \times 100 \times e^{-\frac{-5000}{8.314 \times 298} \times 0.7}=63.63 \text { torr } \\ & p_B=0.7 \times 50 \times e^{-\frac{-5000}{8.314 \times 298} \times 0.3}=66.37 \text { torr } \end{aligned}

Therefore, the correct answer is option A, 63.63 torr and 66.37 torr.


 

Posted by

Deependra Verma

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