Chemistry · The vapour pressure of solutions and Raoult's Law - Ideal and non-ideal solutions, vapour pressure composition, plots for ideal and non-ideal solutions
At the vapour pressure of pure liquid is 520 mmHg and that of pure liquid is 100
At \( 80^{\circ} \mathrm{C}, \) the vapour pressure of pure liquid \( A \) is 520 mmHg and that of pure liquid \( B \) is 1000 mmHg. If a mixture solution of \( A \) and \( B \) boils at \( 80^{\circ} \) and 1 atm pressure, the amount of \( A \) (mole percent) in the mixture is:
- A. \( 50 \% \)
- B. 54\%
- C. \( 32 \% \)
- D. 44\%
Step-by-step solution
Boiling occurs when total vapor pressure equals external pressure (1 atm = 760 mmHg). Using Raoult's law for an ideal solution: P_total = P_A° x_A + P_B° (1 - x_A). Substitute values: 760 = 520x_A + 1000(1 - x_A) => 760 = 1000 - 480x_A => 480x_A = 240 => x_A = 0.5, i.e., 50 mole percent A.
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