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Chemistry
Total vapor pressure of mixture of 1 mol A(pA°=150. torr ) and 2 mol B(pB°=240. torr ) is 200 torr. In this case,
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Q. Total vapor pressure of mixture of $1 \,mol A\left(p_{A}^{\circ}=150\right.$ torr $)$ and $2\, mol B\left(p_{B}^{\circ}=240\right.$ torr $)$ is $200$ torr. In this case,
Solutions
A
there is positive deviation from Raoult's law
B
there is negative deviation from Raoult's law
C
there is no deviation from Raoult's law
D
molecular masses of $A$ and $B$ are also required for calculating the deviation
Solution:
$p_{s}=p_{A}^{\circ} x_{A}+p_{B}^{\circ} x_{B}$
$=\left(150 \times \frac{1}{3}\right)+\left(240 \times \frac{2}{3}\right)=210$
$\Delta p_{\text{mix}}=p_{\exp }-p_{\text {theo }}=- ve$