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[Solved] 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.

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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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Number of moles of liquid A, nA = \(\frac{100}{140}\)mol

= 0.714 mol

Number of moles of liquid B, nB = \(\frac{1000}{180}\)mol

= 5.556 mol

Then, mole fraction of A, xA = \(\frac{n_A}{n_A + n_B}\)

= \(\frac{0.714}{0.714+5.556}\)

= 0.114

And, mole fraction of B, xB = 1 - 0.114

= 0.886

Vapour pressure of pure liquid B, pB0 = 500 torr

Therefore, vapour pressure of liquid B in the solution,

pB = pB0xB

= 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 = pA0xA

⇒ pA0 = \(\frac{p_A}{x_A}\)

= \(\frac{32}{0.114}\)

= 280.7 torr

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

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