JEE MAIN - Chemistry (2019 - 8th April Morning Slot - No. 21)
The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K on
mixing the two liquids, the sum of their initial volume is equal ot the volume of the final mixture.
The mole fraction of liquid B is 0.5 in the mixture, The vapour pressure of the final solution, the
mole fractions of components A and B in vapour phase, respectively are :
500 mmHg. 0.5,0.5
500 mmHg, 0.4, 0.6
450 mmHg, 0.4,0.6
450 mmHg.0.5,0.5
Explanation
Given
$$P_A^0$$ = 400 mm of Hg
$$P_B^0$$ = 600 mm of Hg
Mole fraction of B (xB) in liquid phase = 0.5
Mole fraction of A (xA) in liquid phase = 1 - 0.5 = 0.5
Pressure of solution :
Ps = xA$$P_A^0$$ + xB$$P_B^0$$
= 0.5 $$ \times $$ 400 + 0.5 $$ \times $$ 600
= 500 mm of Hg
From Roult's law we know,
Partial pressure of A (PA) = xA$$P_A^0$$ ......(1)
From Dalton'sRoult's law we know,
Partial pressure of A (PA) = yAPs ........(2)
here yA = mole fraction of A in vapour phase
From (1) and (2) we can write,
xA$$P_A^0$$ = yAPs
$$ \Rightarrow $$ yA = $${{{x_A}P_A^0} \over {{P_s}}}$$ = $${{0.5 \times 400} \over {500}}$$ = 0.4
Mole fraction of A in vapour phase = 0.4
$$ \therefore $$ mole fraction of B in vapour phase = 1 - 0.4 = 0.6
$$P_A^0$$ = 400 mm of Hg
$$P_B^0$$ = 600 mm of Hg
Mole fraction of B (xB) in liquid phase = 0.5
Mole fraction of A (xA) in liquid phase = 1 - 0.5 = 0.5
Pressure of solution :
Ps = xA$$P_A^0$$ + xB$$P_B^0$$
= 0.5 $$ \times $$ 400 + 0.5 $$ \times $$ 600
= 500 mm of Hg
From Roult's law we know,
Partial pressure of A (PA) = xA$$P_A^0$$ ......(1)
From Dalton'sRoult's law we know,
Partial pressure of A (PA) = yAPs ........(2)
here yA = mole fraction of A in vapour phase
From (1) and (2) we can write,
xA$$P_A^0$$ = yAPs
$$ \Rightarrow $$ yA = $${{{x_A}P_A^0} \over {{P_s}}}$$ = $${{0.5 \times 400} \over {500}}$$ = 0.4
Mole fraction of A in vapour phase = 0.4
$$ \therefore $$ mole fraction of B in vapour phase = 1 - 0.4 = 0.6
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