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Karnataka Board PUCPUC Science 2nd PUC Class 12

100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. - Chemistry

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Question

100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Numerical

Solution

Number of moles of liquid A, nA = `100/140  "mol"`

= 0.714 mol

Number of moles of liquid B, nB = `1000/180  "mol"`

= 5.556 mol

Then, the mole fraction of A, xA = `"n"_"A"/("n"_"A"+"n"_"B")`

= `0.714/(0.714+5.556)`

= 0.114

And, mole fraction of B, xB = 1 − 0.114 = 0.886

Vapour pressure of pure liquid B, `"p"_"B"^0` = 500 torr

Therefore, the vapour pressure of liquid B in the solution,

pB = `"p"_"B"^0"x"_"B"`

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, ptotal = 475 torr

∴ Vapour pressure of liquid A in the solution,

pA = ptotal − pB

= 475 − 443

= 32 torr

Now,

pA = `"p"_"A"^0"x"_"A"`

`=>"p"_"A"^0 = "p"_"A"/"x"_"A"`

= `32/0.114`

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

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Chapter 2: Solutions - Exercises [Page 61]

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NCERT Chemistry [English] Class 12
Chapter 2 Solutions
Exercises | Q 36 | Page 61

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