Question

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are :

(1) 1400 mm Hg, A

(2) 1400 mm Hg, B

(3) 600 mm Hg, B

(4) 600 mm Hg, A

Answer 1

Correct option is: (4) 600 mm Hg, A

\(P_{\text{total}} = p^o_A X_A + P_B ^oX_B\)

\(500 = 200 \left( \frac{1}{4} \right) + p_B^o \left( \frac{3}{4} \right)\)

\(2000 = 200 + 3p_B^o\)

\(p_B ^o = 600\) 

\(P_A^o = 200\)

A is least volatile

Answer 2

Correct answer is option (2)

The formula used is p_ax_a + p_bx_b = p_t​​​​

X_a = 1/4

X_b = 3/4 

200×1/4 + 3/4×p_b = 500 

So, p_b = 600 

Less volatile component is A because it's vapour pressure is less