Two volatile liquids have pure vapour pressures p1^0 = 200 mm Hg and p2^0 = 415 mm Hg. If the mole fraction of component 2 in solution is x2 = 0.688, the total vapour pressure of the solution is:
A347.9 mm Hg
B200.0 mm Hg
C285.5 mm Hg
D415.0 mm Hg
Answer & Solution
Correct answer: A. 347.9 mm Hg
1. $p_{total} = p_1^0 + (p_2^0 - p_1^0)x_2$.
2. = $200 + (415 - 200)\times 0.688$.
3. = $200 + 215 \times 0.688 = 200 + 147.9$.
4. = $347.9$ mm Hg.
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.10_
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