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Total vapour pressure of mixture of 1 mole $X(P_x^0=150torr)$ and 2 mol Y$(P_y^0=240torr)$ is 210mm.In this case

$\begin{array}{1 1}(a)\;\text{There is negative deviation from Raoult's law}\\(b)\;\text{There is positive deviation from Raoult's law}\\(c)\;\text{There is no deviation from Raoult's law}\\(d)\;\text{Molecular masses of X & Y are also needed to describe the system}\end{array}$

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  • According to Dalton’s law of partial pressures, the total pressure over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as P=$P_X^0x_X+P_Y^0x_Y$
Answer: There is no deviation from Raoult's law
Given, Moles of X = 1 mole, Moles of Y = 2 moles
Partial pressure of X = $P^0_X \times x_X = 150\; torr \times \frac{1}{1+2} = 150 \times \frac{1}{3} \; torr$
Partial pressure of Y = $P^0_X \times x_X = 240\; torr \times \frac{2}{1+2} = 240 \times \frac{2}{3} \; torr$
So, the total presssure is P=$P_X^0x_X+P_Y^0x_Y$
$\Rightarrow 150\times \large\frac{1}{3}$$+240\times \large\frac{2}{3}$
$\Rightarrow 50+60$
$\Rightarrow 210$ torr
Given total pressure of the solution = 210 mm
Since the solution is obeying Raoult's law. Hence there is no deviation from the Raoult's law.
answered Feb 14, 2014 by sreemathi.v
edited Jul 20, 2014 by mosymeow_1
 

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