At$80^{\circ}C\$, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at $80^{\circ}C$ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

Question

Dr. Abubacker · Accepted Answer

(d)
$P_{T}$=$P_{A}^{\circ}X_{A} + P_{B}^{\circ}x_{B}$
Mixture solution boil at 1 atm = 760 mm = total pressure.
$760 = 520\ X_{A} + 100(1 - X_{A})$
$X_{A} =$ 0.5, mol% of A =50%

Q. At$80^{\circ}C\$, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at $80^{\circ}C$ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

Explanation

Reviewed by Dr. Abubacker

Want more practice?