At 300 K the vapour pressures of two pure liquids. A and B are 100 and 500 mm Hg, respectively. If in a mixture of A and B, the vapour pressure is 300 mm Hg, the mole fractions of A in the liquid an d in the vapour phase, respectively are

Question

At 300 K the vapour pressures of two pure liquids. A and B are 100 and 500 mm Hg, respectively. If in a mixture of A and B, the vapour pressure is 300 mm Hg, the mole fractions of A in the liquid an d in the vapour phase, respectively are (A) 1/2 and 1/10 (B) 1/4 and 1/6 (C) 1/4 and 1/10 (D) 1/2 and 1/6

Tardigrade Admin · Accepted Answer

In liquid phase, according to Daltonâs law of partial pressure PT =PA +PB = xA p°A + xBp°B = xA p°A + (1-xA)p°B PT = (p°A-p°B)xA+p°B 300 = (-400)xA + 500 xA=(1/2) In vapour phase, xA=(pA/pT)=(p°A x A/p°A x A+p°B x B) =(100×(1/2)/100×(1)2+500×(1)2 =(50/300)=(1/6)

Q. At 300K the vapour pressures of two pure liquids. A and B are 100 and 500 mm Hg, respectively. If in a mixture of A and B, the vapour pressure is 300mmHg, the mole fractions of A in the liquid an d in the vapour phase, respectively are

1/2 and 1/10

1/4 and 1/6

1/4 and 1/10

1/2 and 1/6

Q. At $300 K$ the vapour pressures of two pure liquids. $A$ and $B$ are $100$ and $500$ mm $H g$ , respectively. If in a mixture of $A$ and $B$ , the vapour pressure is $300 mm H g$ , the mole fractions of $A$ in the liquid an d in the vapour phase, respectively are