A binary solution contains x1 and x2 mole fraction of two components having vapour pressure p1∘ and p2∘ in their pure states. The total vapour pressure above the solution is

According to Raoult's law, we have p1=p1∘x1 and p2=p2∘x2So, p=p1+p2=p1∘x1+p2∘1−x1∴x1+x2=1p=p1∘−p2∘x1+p2∘

A binary solution contains x1 and x2 mole fraction of two components having vapour pressure p1∘ and p2∘ in their pure states. The total vapour pressure above the solution is

According to Raoult's law, we have p1=p1∘x1 and p2=p2∘x2So, p=p1+p2=p1∘x1+p2∘1−x1∴x1+x2=1p=p1∘−p2∘x1+p2∘

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A binary solution contains x₁and x₂mole fraction of two components having vapour pressure $p_{1}^{\circ} and p_{2}^{\circ}$ in their pure states. The total vapour pressure above the solution is

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$(p_{2}^{\circ} - p_{1}^{\circ}) x_{1} + p_{1}^{\circ}$

$(p_{1}^{\circ} - p_{2}^{\circ}) x_{1} + p_{1}^{\circ}$

$(p_{2}^{\circ} - p_{1}^{\circ}) x_{1} + p_{2}^{\circ}$

$(p_{1}^{\circ} - p_{2}^{\circ}) x_{1} + p_{2}^{\circ}$

answer is A.

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Detailed Solution

According to Raoult's law, we have $p_{1} = p_{1}^{\circ} x_{1} and p_{2} = p_{2}^{\circ} x_{2}$
So,
$\begin{array}{l} p = p_{1} + p_{2} = p_{1}^{\circ} x_{1} + p_{2}^{\circ} (1 - x_{1}) \\ ∴ [x_{1} + x_{2} = 1] \\ p = (p_{1}^{\circ} - p_{2}^{\circ}) x_{1} + p_{2}^{\circ} \end{array}$