At 80- C- the vapour pressure of a pure liquid A is 520 mm Hg and that of a pure liquid of B is 1000 mm Hg- If the ideal mixture of liquids A andB boils at 80- C and 1 atm pressure- the amount of solution A in the mixture is- 1 atm -760 mm Hg A- 50 mol -B- 52 mol -C- 34 mol -D- 48 mol -

The correct option is A 50 mol %We know that, nbsp; P_T = P^o_A x_A+ P^o_B x_B x_A and x_B mole fraction of A and B 760 = 520 x_A +(1000 (1 x_A) x_A = 1/2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Vapour Pressure

At 80∘ C, the...

Question

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

50 mol %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

52 mol %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

34 mol %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

48 mol %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $50 mol %$
We know that,
$P_{T} = P_{A}^{o} x_{A} + P_{B}^{o} x_{B}$
$x_{A} a n d x_{B}$ mole fraction of A and B
$760 = 520 x_{A} + (1000 (1 - x_{A})$
$x_{A} = \frac{1}{2} = 0.5$
So, 50 mol % of A is present in the mixture.

Suggest Corrections

Similar questions

Q. At

80^{o} C

, the vapour pressure of pure liquid

A

' is

520 m m H g

and that of pure liquid

B

' is

1000 m m H g

. If a mixture solution of

A

' and

B

' boils at

80^{o} C

and 1 atm pressure, the amount of

A

' in the mixture is:

[

1 a t m = 760 m m H g

]

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 a pure liquid

A

520 mm Hg

and that of a pure liquid of

B

1000 mm Hg

. If the ideal mixture of liquids

A

and

B

boils at

80^{\circ} C

and 1 atm pressure, the amount of solution A in the mixture is: (1 atm = 760 mm Hg)

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 mixture of both liquids from ideal solution and 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:

Join BYJU'S Learning Program

At 80∘ C, the vapour pressure of a pure liquid A is 520 mm Hg and that of a pure liquid of B is 1000 mm Hg. If the ideal mixture of liquids A and B boils at 80∘ C and 1 atm pressure, the amount of solution A in the mixture is: (1 atm = 760 mm Hg)

The correct option is A 50 mol %We know that, PT=PoAxA+PoBxB xA and xB mole fraction of A and B 760=520xA+(1000(1−xA) xA=12=0.5 So, 50 mol % of A is present in the mixture.

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

The correct option is A $50 mol %$
We know that,
$P_{T} = P_{A}^{o} x_{A} + P_{B}^{o} x_{B}$
$x_{A} a n d x_{B}$ mole fraction of A and B
$760 = 520 x_{A} + (1000 (1 - x_{A})$
$x_{A} = \frac{1}{2} = 0.5$
So, 50 mol % of A is present in the mixture.