[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.
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.
Number of moles of liquid A, n_{A} = \(\frac{100}{140}\)mol
= 0.714 mol
Number of moles of liquid B, n_{B} = \(\frac{1000}{180}\)mol
= 5.556 mol
Then, mole fraction of A, x_{A} = \(\frac{n_A}{n_A + n_B}\)
= \(\frac{0.714}{0.714+5.556}\)
= 0.114
And, mole fraction of B, x_{B} = 1  0.114
= 0.886
Vapour pressure of pure liquid B, p_{B}^{0} = 500 torr
Therefore, vapour pressure of liquid B in the solution,
p_{B} = p_{B}^{0}x_{B}
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution, p_{total} = 475 torr
∴ Vapour pressure of liquid A in the solution,
p_{A} = p_{total}  p_{B}
= 475  443
= 32 torr
Now, p_{A} = p_{A}^{0}x_{A}
⇒ p_{A}^{0} = \(\frac{p_A}{x_A}\)
= \(\frac{32}{0.114}\)
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.

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