Q1 / 75 · JEE Main 2026 (28 January Shift 1) · Chemistry
At $\mathrm{T}(\mathrm{K}), 2$ moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:
- A400, 300
- B600,400
- C300,200
- Correct answerD500, 200
Solution
For an ideal solution obeying Raoult's Law, total pressure is
$P = P_A^o x_A + P_B^o x_B$.
Initially with 2 moles A and 3 moles B:
$x_A = 0.4, x_B = 0.6$, and $P = 320 = 0.4P_A^o + 0.6P_B^o$
After adding 1 mole each: 3 moles A and 4 moles B gives
$x_A = \frac{3}{7}, x_B = \frac{4}{7}$, and $P = 328.6 = \frac{3}{7}P_A^o + \frac{4}{7}P_B^o$
From the first equation:
$3200 = 4P_A^o + 6P_B^o$
From the second:
$2300.2 = 3P_A^o + 4P_B^o$
Solving: We find $P_B^o = 200$ mm Hg and $P_A^o = 500$ mm Hg.
$P = P_A^o x_A + P_B^o x_B$.
Initially with 2 moles A and 3 moles B:
$x_A = 0.4, x_B = 0.6$, and $P = 320 = 0.4P_A^o + 0.6P_B^o$
After adding 1 mole each: 3 moles A and 4 moles B gives
$x_A = \frac{3}{7}, x_B = \frac{4}{7}$, and $P = 328.6 = \frac{3}{7}P_A^o + \frac{4}{7}P_B^o$
From the first equation:
$3200 = 4P_A^o + 6P_B^o$
From the second:
$2300.2 = 3P_A^o + 4P_B^o$
Solving: We find $P_B^o = 200$ mm Hg and $P_A^o = 500$ mm Hg.
Hence, the correct answer is option: 'D'.