JEE Chemistry Trick: Vapour Phase Composition from Raoult’s Law Explained 💡

 

❓ Question

Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If xA and xB are the mole fraction of A and B in solution while yA and yB are the mole fraction of A and B in vapour phase then

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🖼️ Question Image

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✍️ Short Solution

For an ideal solution Raoult’s law applies:

$$p_A \;=\; x_A\,p_A^*,\qquad p_B \;=\; x_B\,p_B^*$$

Total vapour pressure above the solution:

$$p_{\text{tot}} \;=\; p_A + p_B \;=\; x_A p_A^* + x_B p_B^*.$$

The mole fractions in the vapour (via Dalton’s law):

$$y_A \;=\; \frac{p_A}{p_{\text{tot}}} =\frac{x_A p_A^*}{x_A p_A^* + x_B p_B^*},$$
$$y_B \;=\; \frac{p_B}{p_{\text{tot}}} =\frac{x_B p_B^*}{x_A p_A^* + x_B p_B^*}.$$

Since $x_B=1-x_A$, you can write $y_A$ solely in terms of $x_A$:

$$\boxed{\,y_A=\dfrac{350\,x_A}{350\,x_A+750(1-x_A)}=\dfrac{350x_A}{750-400x_A}\,}$$

and similarly

$$\boxed{\,y_B=\dfrac{750(1-x_A)}{750-400x_A}\,}.$$

These give the vapour composition for any liquid composition $x_A$.

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🧮 Some useful special cases

• If $x_A=0$ (pure B), then $y_A=0$ (no A in vapour).

• If $x_A=1$ (pure A), then $y_A=1$.

• For an equimolar liquid $x_A=0.5$:

$$y_A=\frac{350\times0.5}{350\times0.5+750\times0.5}=\frac{175}{175+375}=\frac{175}{550}\approx0.318.$$

So the vapour is richer in B (since B is more volatile).

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🖼️ Image Solution

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✅ Conclusion & Physical Insight

• Key formulas (Raoult + Dalton) give the direct mapping between liquid and vapour compositions:

  $$y_A=\dfrac{x_A p_A^*}{x_A p_A^* + x_B p_B^*},\quad y_B=1-y_A.$$
  

• Because $p_B^*(750) > p_A^*(350)$, B is more volatile; for a given liquid mixture the vapour will always be richer in B than the liquid (i.e. $y_B>x_B$ for typical compositions).

• Use the simplified forms above to compute any required numerical $y$ for a given $x$.