Multiple choice

At $80^{\mathrm{o}}\mathrm{C}$, the vapour pressure of pure liquid $\mathrm{A}$' is $520\ mm\ Hg$ and that of pure liquid $\mathrm{B}$' is $1000\ mm\ Hg$. If a mixture solution of $\mathrm{A}$' and $\mathrm{B}$' boils at $80^{\mathrm{o}}\mathrm{C}$ and 1 atm pressure, the amount of $\mathrm{A}$' in the mixture is: [$1\ atm=760\ mm\ Hg$]

  1. $48\ mol\%$
  2. $50\ mol\% $
  3. $52\ mol\%$
  4. $34\ mol\% $
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B Correct answer
Explanation

Raoult's Law: P_total = xA*PA + xB*PB. 760 = xA*520 + (1-xA)*1000. 760 = 520xA + 1000 - 1000xA. 480xA = 240. xA = 240/480 = 0.5 = 50%.

AI explanation

By Raoult's Law, the total vapour pressure of the solution is the sum of the partial pressures of its components, expressed as 760 = 520 multiplied by the mole fraction of A plus 1000 multiplied by the mole fraction of B. Letting the mole fraction of A be x makes the mole fraction of B equal to 1 minus x, which establishes the equation 760 = 520x + 1000(1 minus x). Solving for x gives x = 0.5, meaning liquid A constitutes 50 mol% of the mixture.

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