QUESTION 2.36

100 g of liquid A (molar mass $140 g m o l^{- 1}$ ) was dissolved in 1000 g of liquid B (molar mass $180 g m o l^{- 1}$ ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

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Solution

Step I: Calculation of vapour pressure of pure liquid $A (p_{A}^{\circ})$
Number of moles of liquid A
$(n_{A}) = \frac{W_{A}}{M_{A}} = \frac{(100 g)}{(140 g m o l^{- 1})} = 0.7143 m o l$
Number of moles of liquid B
$(n_{B}) = \frac{W_{B}}{M_{B}} = \frac{(1000 g)}{(180 g m o l^{- 1})} = 5.5556 m o l$
Mole fraction of A
$(x_{A}) = \frac{n_{A}}{n_{A} + n_{B}}$
$= \frac{(0.7143 m o l)}{(0.7143 + 5.5556) m o l} = \frac{0.7143}{6.2699} = 0.1139$
Mole fraction of $B (x_{B}) = 1 - 0.1139 = 0.8861$
Vapour pressure of pure liquid $B (p_{B}^{\circ}) = 500 t o r r$
Total vapour pressure of solution (p) = 475 torr
According to the Raoult's law
$p = p_{A}^{\circ} x_{A} + p_{B}^{\circ} x_{B} 475 t o r r = p_{A}^{\circ} \times (0.1139) + 500 t o r r \times (0.8861) 475 t o r r = p_{A}^{\circ} \times (0.1139) + 443.05 t o r r p_{A}^{\circ} = \frac{(475 - 443.05) t o r r}{(0.1139)} = \frac{31.95}{0.1139} t o r r = 280.5 t o r r$
Step II: Calculation of vapour pressure of A in the solution (p_A)
According to Raoult's law,
$p_{A} = p_{A}^{\circ} x_{A} = (280.5 t o r r) \times (0.1139)$
$p_{A} = 32.0 t o r r$

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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.