JEE Advance - Chemistry (2020 - Paper 2 Offline - No. 14)
Liquids A and B form ideal solution for all compositions of A and B at 25$$^\circ $$C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapour pressure of 0.3 and 0.4 bar, respectively. What is the vapour pressure of pure liquid B in bar?
Answer
0.2
Explanation
Using Raoult's law equation for a mixture of volatile liquids.
$${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$$
$$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$$ ....(i)
$$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$$ ....(ii)
By solving equation (i) and (ii)
$$p_A^o$$ = 0.6 bar and $$p_B^o$$ = 0.2 bar
Thus, the vapour pressure of pure liquid B in bar is 0.2.
$${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$$
$$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$$ ....(i)
$$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$$ ....(ii)
By solving equation (i) and (ii)
$$p_A^o$$ = 0.6 bar and $$p_B^o$$ = 0.2 bar
Thus, the vapour pressure of pure liquid B in bar is 0.2.
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