At 300 K, two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg, In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is

Question

At 300 K, two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg, In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is (A) 0.6 (B) 0.5 (C) 0.8 (D) 0.4

Tardigrade Admin · Accepted Answer

Let the moles of A=x ∴ The moles of B=x[∵ mixture is equimolar.] Mole fraction of A=(x/x+x)=0.5= mole fraction of B. Thus, pT=pAo XA+pBo XB pT=150 × 0.5+100 × 0.5 =125 ∴ Mole fraction of B in vapour mixture =(pBo XB/pT) =(100 × 0.5/125)

Q. At 300K, two pure liquids A and B have vapour pressures respectively 150mmHg and 100mmHg, In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is

0.6

0.5

0.8

0.4

Let the moles of A=x ∴ The moles of B=x[∵ mixture is equimolar.] Mole fraction of A=x+xx​=0.5= mole fraction of B. Thus, pT​=pAo​XA​+pBo​XB​ pT​=150×0.5+100×0.5 =125 ∴ Mole fraction of B in vapour mixture =pT​pBo​XB​​ =125100×0.5​

Q. At $300 K$ , two pure liquids $A$ and $B$ have vapour pressures respectively $150 mm H g$ and $100 mm H g$ , In an equimolar liquid mixture of $A$ and $B,$ the mole fraction of $B$ in the vapour mixture at this temperature is