JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
At T(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:
- A 300, 200
- B 600, 400
- C 400, 300
- D 500, 200
Answer & Solution
Correct Answer
(D) 500, 200
Step-by-step Solution
Detailed explanation
\(P_S=X_A P_A^{\circ}+X_B P_B^{\circ}\) \(320= P _{ A }^{\circ}\left(\frac{2}{5}\right)+ P _{ B }^{\circ}\left(\frac{3}{5}\right)\) \(2 P_{A}^{\circ}+3 P_{B}^{\circ}=1600\) ...(I) Now 1 mole of A & 1 mole of B is added…
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