If vapour pressures of pure liquids 'A' ‘B’ are 300 and 800 torr respectively at 25°C. When these two liquids are mixed at this temperature to form a solution in which mole percentage of ‘B’ is 92, then the total vapour pressure is observed to be 0.95 atm. Which of the following is true for this solution

Question

If vapour pressures of pure liquids 'A' ‘B’ are 300 and 800 torr respectively at 25°C. When these two liquids are mixed at this temperature to form a solution in which mole percentage of ‘B’ is 92, then the total vapour pressure is observed to be 0.95 atm. Which of the following is true for this solution (A) Δ V textmix > 0 (B) Δ H textmix< 0 (C) Δ V textmix=0 (D) Δ H textmix=0

Tardigrade Admin · Accepted Answer

According to Raoult’s law PT=(0.08 × 300+0.92× 800) torr = (24 + 736) torr =760 torr =1 atm Pexp = 0.95 atm < 1 atm Hence solution shows -ve deviation So Δ H textmix < 0 and Δ V textmix < 0

$Δ V_{mix} > 0$

$Δ H_{mix} < 0$

$Δ V_{mix} = 0$

$Δ H_{mix} = 0$

According to Raoult’s law
$P_{T} = (0.08 \times 300 + 0.92 \times 800)$ torr
$= (24 + 736)$ torr
$= 760$ torr $= 1$ atm
$P_{e x p} = 0.95 a t m < 1 a t m$
Hence solution shows -ve deviation
So $Δ H_{mix} < 0$ and $Δ V_{mix} < 0$

ΔVmix​>0

ΔHmix​<0

ΔVmix​=0

ΔHmix​=0

According to Raoult’s law PT​=(0.08×300+0.92×800) torr =(24+736) torr =760 torr =1 atm Pexp​=0.95atm<1atm Hence solution shows -ve deviation So ΔHmix​<0 and ΔVmix​<0

$Δ V_{mix} > 0$

$Δ H_{mix} < 0$

$Δ V_{mix} = 0$

$Δ H_{mix} = 0$

According to Raoult’s law
$P_{T} = (0.08 \times 300 + 0.92 \times 800)$ torr
$= (24 + 736)$ torr
$= 760$ torr $= 1$ atm
$P_{e x p} = 0.95 a t m < 1 a t m$
Hence solution shows -ve deviation
So $Δ H_{mix} < 0$ and $Δ V_{mix} < 0$