Since, PA0=7×103Pa;PB0=12×103Pa and XA=0.4, XB=1−0.4=0.6
According to Raoult's Law: PT=PA0XA+PB0XB
PT=(7×103)(0.4)+(12×103)(0.6)
PT=10×103Pa
Now, PA0XA=PTxA
xA=PTPA0XA=1047×103×0.4⇒0.28
xB=1−xA⇒1−0.28⇒0.72
JEE Main 2019 — Chemistry Physical Chemistry
Liquids A and B form an ideal solution in the entire composition range. At 350K, the vapour pressure of pure A and pure B are 7×103Pa and 12×103Pa , respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
xA=0.4;xB=0.6
xA=0.76;xB=0.24
xA=0.28;xB=0.72
xA=0.37;xB=0.63
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