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प्रश्न
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.
उत्तर
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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