Given PA0=21kPa⇒PB0=18kPa
→ An Ideal solution is prepared by mixing 1 mole A and 2 mole B.
→XA=31 and XB=32
→ According to Raoult's law
PT=XAPA0+XBPB0
⇒PT=(31×21)+(32×18)
⇒PT=7+12=19kPa
JEE Main 2021 — Chemistry Physical Chemistry
At 363K, the vapour pressure of A is 21kPa and that of B is 18kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is___________ kPa. (Round of to the Nearest Integer).
Held on 16 Mar 2021 · Verified 6 Jul 2026.
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