The vapour pressure of a solution of the liquids A P0-80 mm- of Hg and -X A -0-4 and B P 0-120 mm- of Hg and -X B -0-6 is observed to be 100 mm of Hg- It shows that the solution exhibits-A- Positive deviation from ideal behaviourB- Negative deviation from ideal behaviourC- Ideal behaviourD- Positive deviation at low concentrations and negative at higher concentrations

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Standard XII

Chemistry

Azeotropic Mixtures

The vapour pr...

Question

The vapour pressure of a solution of the liquids A $(P^{0} = 80 m m o f H g a n d X_{A} = 0.4)$ and B $(P^{0} = 120 m m o f H g a n d X_{B} = 0.6)$ is observed to be 100 mm of Hg. It shows that the solution exhibits:

Positive deviation from ideal behaviour

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Negative deviation from ideal behaviour

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Ideal behaviour

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Positive deviation at low concentrations and negative at higher concentrations

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Solution

The correct option is B Negative deviation from ideal behaviour
From Raoults law,
$P_{t o t a l} = p_{A}^{0} \times X_{A} + p_{B}^{0} \times X_{B}$
$= 80.0 \times 0.4 + 120.0 \times 0.6$
$= 104 m m o f H g$
The observed $P_{t o t a l}$ is 100 mm of Hg which is less than 104 mm of Hg. Hence the solution shows negative deviation.

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Similar questions

The vapour pressure of the solution of two liquids $A (p^{\circ} = 80 m m)$ and $B (p^{\circ} = 120 m m)$ is found to be 100 mm when $X_{A}$ = 0.4. The result shows that

Q. The vapour pressure of the solution of two liquids A

(p^{\circ} = 80 m m)

and

(p^{\circ} = 120 m m)

is found to be

100 m m

when

χ_{A} = 0.4

. The result shows that:

Q. Vapour pressures of pure acetone and chloroform at

328 K

are

741.8 m m H g

and

632.8 m m H g

respectively. Assuming that they form ideal solution over the entire range of composition, plot

p_{t o t a l}, p_{c h l o r o f o r m}

, and

p_{a c e t o n e}

as a function of

x_{a c e t o n e}

. The experimental data observed for different compositions of mixture is:

$100 x x_{a c e t o n e}$	$0$	$11.8$	$23.4$	$36.0$	$50.8$	$58.2$	$64.5$	$72.1$
$P_{a c e t o n e} / m m H g$	$0$	$54.9$	$110.1$	$202.4$	$322.7$	$405.9$	$454.1$	$521.1$
$P_{c h l o r o f o r m} / m m H g$	$632.8$	$548.1$	$469.4$	$359.7$	$257.7$	$193.6$	$161.2$	$120.7$

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Q. CASE1: Read the passage given below and answer the following question.

The solution which obeys Raoult’s law over the entire range of concentration are known as ideal solutions. The solution which does not obey Raoults law over the entire range of concentration, then it is called non-ideal solution. The vapour pressure of such as a solution is either higher or lower than that predicted by Raoult’s law. If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation from Raoult’s law.
Some liquids on mixing, form azeotropes which are binary mixture having the same composition in liquid and vapour phase. There are two types of azeotropes. Minimum boiling azeotropes and maximum boiling azeotropes.

Which of the following solutions show negative deviation from Raoult’s law?
[0.77 Mark]

QUESTION 2.37

Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot $p_{t o t a l}$ , $p_{c h l o r o f o r m}$ and $p_{a c e t o n e}$ . The experimental data observed for different compoisition of mixture is:

$\begin{matrix} 100 x_{a c e t o n e} & 0 & 11.8 & 23.4 & 36.0 & 50.8 & 58.2 & 64.5 & 72.1 p_{a c e t o n e} / m m H g & 0 & 54.9 & 110.1 & 202.4 & 322.7 & 405.9 & 454.1 & 521.1 p_{c h l o r o f o r m} / m m H g & 632.8 & 548.1 & 469.4 & 359.7 & 257.7 & 193.6 & 161.2 & 120.7 \end{matrix}$

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

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Azeotropic Mixtures

Standard XII Chemistry

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The vapour pressure of a solution of the liquids A (P0=80 mm of Hg and XA=0.4) and B (P0=120 mm of Hg and XB=0.6) is observed to be 100 mm of Hg. It shows that the solution exhibits:

The correct option is B Negative deviation from ideal behaviourFrom Raoults law, Ptotal=p0A×XA+p0B×XB =80.0×0.4+120.0×0.6 =104 mm of Hg The observed Ptotal is 100 mm of Hg which is less than 104 mm of Hg. Hence the solution shows negative deviation.

The vapour pressure of a solution of the liquids A $(P^{0} = 80 m m o f H g a n d X_{A} = 0.4)$ and B $(P^{0} = 120 m m o f H g a n d X_{B} = 0.6)$ is observed to be 100 mm of Hg. It shows that the solution exhibits: