Single Choice

Two moles of pure liquid 'A' $$(P_{A}^{0} =80mm$$ of Hg) and $$3$$ moles of pure liquid 'B' ($$P_{B}^{0} = 120mm$$ of Hg) are mixed. Assuming ideal behaviour?

AVapour pressure of the mixture is $$104$$mm of Hg
Correct Answer
BMole fraction of liquid 'A' in Vapour pressure is $$0.3077$$
CMole fraction of 'B' in Vapour pressure is $$0.692$$
DMole fraction of 'B' in Vapour pressure is $$0.785$$

Solution

$$P_{A}^{\circ}=80 mm Hg$$ $$P_{B}^{\circ}=120 mm Hg$$ $$n_{A}=2 moles$$ $$n_{B}=3 moles$$ $$x_{A}=\dfrac{2}{2+3}=\dfrac{2}{5}=0.4$$ $$x_{B}=\dfrac{3}{5}=0.6$$ $$P_{A}=P_{A}^{\circ}x_{A}=80\times 0.4=32 mm Hg$$ $$P_{B}=P_{B}^{\circ}x_{B}=120\times 0.6=72 mm Hg$$ Total vapour pressure $$P=P_{A}+P_{B}$$ $$=32+72=104 mm Hg$$

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