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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 1
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
And, mole fraction of B, χB = 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 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.
Solution 2
Number of moles of liquid A = `100/140`
= 0.714
Number of moles of liquid B = `1000/180`
= 5.555
Hence, mole fraction of A in the solution
`chi_A = 0.714/(0.714 + 5.555)`
= 0.114
and mole fraction of B in the solution
`chi_B = 5.555/(0.714 + 5.555)`
= 0.886
According to Raoult’s law,
`p_A = chi_Ap_A^circ`
`p_A = p_A^circ` ...(i)
`p_A = chi_Bp_B^circ`
= 0.886 × 500
= 443 torr ...(ii)
∴ Total vapour pressure of the solution
p = pA + pB
= `0.114 p_A^circ + 443`
Given that p = 475 torr
∴ `0.114 p_A^circ + 443 = 475`
or `p_A^circ = (475 - 443)/0.114`
= 280.7 torr
Putting this value in eq. (i), we have
pA = 0.114 × 280.7
= 31.9998
= 32 torr
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