At $80^{\circ}$ C the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liq ‘B’ is 1000 mm Hg. If a mixture of both the liquids form an ideal solution which boils at $80^{\circ} C$ and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

Q. At

80^{\circ}

C the vapour pressure of pure liquid ‘A’ is 520 mm of Hg and that of pure liq ‘B’ is 1000 mm of Hg. If the mixture of both liquids from an ideal solution and boils at

80^{\circ}

C and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

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At 800C, the vapour pressure of pure liquid A is 250mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at 800C and 1 atm pressure, the amount of A in the mixture is :(1atm=760mmHg)

The correct option is B 32 mole percent1 atm = 760 mm Hg = PTPT=P0AXA×P0BXB760=250XA+1000(1−XA)240=750XAXA=0.32Hence mole % of A = 32%

At $80^{0}$ C, the vapour pressure of pure liquid A is $250$ mm of Hg and that of pure liquid B is $1000$ mm of Hg. If a solution of A and B boils at $80^{0}$ C and $1$ atm pressure, the amount of A in the mixture is : $(1 a t m = 760 m m$ Hg)

The correct option is B $32$ mole percent
1 atm = 760 mm Hg = $P_{T}$
$P_{T} = P_{A}^{0} X_{A} \times P_{B}^{0} X_{B}$
$760 = 250 X_{A} + 1000 (1 - X_{A})$
$240 = 750 X_{A}$
$X_{A} = 0.32$
Hence mole $%$ of A = $32 %$