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Standard XII

Physics

Gas laws from KTG

At 40oC, th...

Question

At $40^{o} C$ , the vapour pressure of pure liquids, benzene and toluene, are 160 mm $H g$ and 60 mm $H g$ , respectively. At the same temperature, the vapor pressure of an equimolar solution of the two liquids, assuming the ideal solution, should be:

140 mm

H g

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110 mm

H g

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220 mm

H g

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100 mm

H g

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Solution

The correct option is C 110 mm $H g$
As we know,

$P_{m} = P_{1}^{o} x_{1} + P_{2}^{o} x_{2}$

equimolar means $x_{1} = x_{2} = 0.5$

$P_{m} = 160 \times \frac{1}{2} + 60 \times \frac{1}{2} = 110 m m$

Hence, the correct option is $B$

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Similar questions

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(in mm Hg upto two decimal).

Q. The vapour pressure of pure liquids 1 and 2 are

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Benzene and toluene form nearly ideal solution .At 313K the vapour pressure of pure benzene is 150mm Hg and of pure toluene is 50 mm Hg .calculate vapour pressure of mixture of these two containing their equal masses at 313 K .

Q. The vapour pressure of pure liquids 1 and 2 are

450 m m H g

and

700 m m H g

respectively at

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. At a point when total vapour pressure is

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, choose the correct option(s) :

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At 40oC, the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg, respectively. At the same temperature, the vapor pressure of an equimolar solution of the two liquids, assuming the ideal solution, should be:

The correct option is C 110 mm HgAs we know,Pm=Po1x1+Po2x2equimolar means x1=x2=0.5Pm=160×12+60×12=110 mmHence, the correct option is B

At $40^{o} C$ , the vapour pressure of pure liquids, benzene and toluene, are 160 mm $H g$ and 60 mm $H g$ , respectively. At the same temperature, the vapor pressure of an equimolar solution of the two liquids, assuming the ideal solution, should be:

The correct option is C 110 mm $H g$
As we know,

$P_{m} = P_{1}^{o} x_{1} + P_{2}^{o} x_{2}$

equimolar means $x_{1} = x_{2} = 0.5$

$P_{m} = 160 \times \frac{1}{2} + 60 \times \frac{1}{2} = 110 m m$

Hence, the correct option is $B$