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Standard XII

Chemistry

Raoult's Law

At 80o C, t...

Question

At $80^{o}$ C, the vapour pressure of pure liquid $A$ is 520 mmHg and that of pure liquid $B$ is 1000 mmHg. If a mixture solution of $A$ and $B$ boils at $80^{o}$ and 1 atm pressure, the amount of $A$ (mole percent) in the mixture is:

50

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Solution

The correct option is B $50$ %
Let the amount of $A$ in the mixture be $χ_{A}$ and B be the $χ_{B}$ .

$P_{T} = 760 = P_{A}^{o} χ_{A} + P_{B}^{o} χ_{B}$

$760 = 520 χ_{A} + 1000 (1 - χ_{A})$

$480 χ_{A} = 240$

$χ_{A} = \frac{1}{2}$ or $50 %$ .

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At 80o C, the vapour pressure of pure liquid A is 520 mmHg and that of pure liquid B is 1000 mmHg. If a mixture solution of A and B boils at 80o and 1 atm pressure, the amount of A (mole percent) in the mixture is:

The correct option is B 50%Let the amount of A in the mixture be χA and B be the χB.PT=760=PoAχA+PoBχB760=520χA+1000(1−χA)480χA=240χA=12 or 50%.

At $80^{o}$ C, the vapour pressure of pure liquid $A$ is 520 mmHg and that of pure liquid $B$ is 1000 mmHg. If a mixture solution of $A$ and $B$ boils at $80^{o}$ and 1 atm pressure, the amount of $A$ (mole percent) in the mixture is:

The correct option is B $50$ %
Let the amount of $A$ in the mixture be $χ_{A}$ and B be the $χ_{B}$ .

$P_{T} = 760 = P_{A}^{o} χ_{A} + P_{B}^{o} χ_{B}$

$760 = 520 χ_{A} + 1000 (1 - χ_{A})$

$480 χ_{A} = 240$

$χ_{A} = \frac{1}{2}$ or $50 %$ .