The total pressure observed by mixing two liquids A and B is 350 mm Hg when their mole fractions are 0.7 and 0.3 respectively.
The total pressure becomes 410 mm Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is __________ mm Hg. (Nearest integer)
Consider the liquids and solutions behave ideally.

Solution

Chemistry ma’am · 4 steps

Two mixtures give two Raoult's law equations, and two equations are enough to find both pure vapour pressures.

1. Raoult's law. For an ideal solution, the total pressure is the sum of each liquid's share: P = x_A p_A° + x_B p_B°. Each mixture gives one equation in the two unknowns p_A° and p_B°.

   0.7 p_A° + 0.3 p_B° = 350 ; 0.2 p_A° + 0.8 p_B° = 410

2. Eliminate pB. Multiply the first equation by 8 to get 5.6 p_A° + 2.4 p_B° = 2800. Multiply the second by 3 to get 0.6 p_A° + 2.4 p_B° = 1230. Subtracting leaves 5 p_A° = 1570.
3. Check. p_A° = 314 mm Hg. Putting this back gives p_B° = (350 - 219.8)/0.3 = 434 mm Hg, and 0.2 × 314 + 0.8 × 434 = 410 confirms it.
4. So. The vapour pressure of pure A is 314 mm Hg.

The common slip: 434. Solving correctly but reporting the pure vapour pressure of B gives 434 instead of the value for A.

The idea to keep. Whenever two compositions and two total pressures are given, write P = x_A p_A° + x_B p_B° for each and solve the pair. Always check which unknown the question asks for.

Answer from the NTA final answer key. Solution by PrepWiser.