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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.
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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 (χA) = `n_A/(n_A + n_B)`
= `0.714/(0.714 + 5.556)`
= 0.114
Mole fraction of B (χB) = 1 − 0.114
= 0.886
Vapour pressure of pure liquid B `(p_B^0)` = 500 torr
∴ The vapour pressure of liquid B in the solution,
pB = `p_B^0 chi_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 chi_A`
⇒ `p_A^0 = p_A/chi_A`
= `32/0.114`
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
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