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:

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At 80oC, 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 80o and 1 atm pressure, the amount of A in the mixture is? (1 atm =760 mm Hg).

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

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

The correct option is C $32$ 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 $= 32 %$