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,

Question

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, (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

Tardigrade Admin · Accepted Answer

ps=pA° xA+pB° xB =(150 × (1/3))+(240 × (2/3))=210 Δ p textmix=p exp -p text theo =- ve

Q. Total vapor pressure of mixture of $1 m o l A (p_{A}^{\circ} = 150$ torr $)$ and $2 m o lB (p_{B}^{\circ} = 240$ torr $)$ is $200$ torr. In this case,

there is positive deviation from Raoult's law

there is negative deviation from Raoult's law

there is no deviation from Raoult's law

molecular masses of $A$ and $B$ are also required for calculating the deviation

$p_{s} = p_{A}^{\circ} x_{A} + p_{B}^{\circ} x_{B}$

$= (150 \times \frac{1}{3}) + (240 \times \frac{2}{3}) = 210$

$Δ p_{mix} = p_{e x p} - p_{theo} = - v e$

Q. Total vapor pressure of mixture of 1molA(pA∘​=150 torr ) and 2molB(pB∘​=240 torr ) is 200 torr. In this case,