Consider a binary mixture of volatile liquids. If at XA = 0.4 the vapour pressure of solution is 580 torr then the mixture could be (pA°=300 torr, pB°=800 torr) :

Question

Consider a binary mixture of volatile liquids. If at XA = 0.4 the vapour pressure of solution is 580 torr then the mixture could be (pA°=300 torr, pB°=800 torr) : (A) CHCl3 - CH3COCH3 (B) C6 H5 Cl - C6 H4Br (C) C6H6-C6H5CH3 (D) nC6 H14-n C7 H16

Tardigrade Admin · Accepted Answer

It shows negative deviation from Raoult’s law ps (actual) = 580 torr ps (Raoult) = 0.4 × 300+0.6× 800 = 600 torr

Q. Consider a binary mixture of volatile liquids. If at $X_{A} = 0.4$ the vapour pressure of solution is 580 torr then the mixture could be $(p_{A}^{\circ} = 300$ torr, $p_{B}^{\circ} = 800$ torr) :

$C H C l_{3} - C H_{3} COC H_{3}$

$C_{6} H_{5} Cl - C_{6} H_{4} B r$

$C_{6} H_{6} - C_{6} H_{5} C H_{3}$

$n C_{6} H_{14} - n C_{7} H_{16}$

It shows negative deviation from Raoult’s law
$p_{s}$ (actual) = 580 torr
$p_{s}$ (Raoult) $= 0.4 \times 300 + 0.6 \times 800 = 600$ torr

Q. Consider a binary mixture of volatile liquids. If at XA​=0.4 the vapour pressure of solution is 580 torr then the mixture could be (pA∘​=300 torr, pB∘​=800 torr) :

CHCl3​−CH3​COCH3​

C6​H5​Cl−C6​H4​Br

C6​H6​−C6​H5​CH3​

nC6​H14​−nC7​H16​