Question

Two volatile liquids X and Y form an ideal solution at 298 K and have vapour pressures equal to 100 mm and 200 mm of Hg respectively in their pure state. The mole fraction of X in the solution is 0.4 and the mole fraction of Y in the vapour phase is \( \frac{a}{20} \). Calculate the value of \(a\).

• A. 25
• B. 15
• C. 10
• D. 5

Hint:

For ideal solutions, use Raoult's law \(P_i=x_iP_i^0\), and for vapour phase mole fraction use \(y_i=\frac{P_i}{P_{\text{total}}}\).

Answer 1

The Correct Option is B

Step 1: Write the given data.
Pure vapour pressure of liquid X is:
\[ P_X^0 = 100\ \text{mm Hg} \]
Pure vapour pressure of liquid Y is:
\[ P_Y^0 = 200\ \text{mm Hg} \]
Mole fraction of X in solution is:
\[ x_X = 0.4 \]

Step 2: Find mole fraction of Y in solution.
Since the solution contains only X and Y:
\[ x_X+x_Y=1 \]
\[ x_Y=1-0.4=0.6 \]

Step 3: Apply Raoult's law for X.
Partial vapour pressure of X is:
\[ P_X=x_XP_X^0 \]
\[ P_X=0.4 \times 100=40\ \text{mm Hg} \]

Step 4: Apply Raoult's law for Y.
Partial vapour pressure of Y is:
\[ P_Y=x_YP_Y^0 \]
\[ P_Y=0.6 \times 200=120\ \text{mm Hg} \]

Step 5: Find total vapour pressure.
\[ P_{\text{total}}=P_X+P_Y \]
\[ P_{\text{total}}=40+120=160\ \text{mm Hg} \]

Step 6: Find mole fraction of Y in vapour phase.
Mole fraction of Y in vapour phase is:
\[ y_Y=\frac{P_Y}{P_{\text{total}}} \]
\[ y_Y=\frac{120}{160} \]
\[ y_Y=\frac{3}{4} \]

Step 7: Compare with given form.
Given mole fraction of Y in vapour phase is:
\[ \frac{a}{20} \]
So,
\[ \frac{a}{20}=\frac{3}{4} \]
\[ a=20 \times \frac{3}{4} \]
\[ a=15 \]
Final Answer:
\[ \boxed{15} \]