The vapour pressures of ethanol and methanol are 44.5 mm Hg and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. The total vapour pressure of the solution and the mole fraction of methanol in the vapour are respectively

Question

The vapour pressures of ethanol and methanol are 44.5 mm Hg and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. The total vapour pressure of the solution and the mole fraction of methanol in the vapour are respectively (A) 43.46 mm and 0.51 (B) 66.15 mm and 0.657 (C) 66.15 mm and 0.791 (D) 70.59 mm and 0.657

Tardigrade Admin · Accepted Answer

We know, PTotaI = p1 + p2 Also p1 = p°1 × x1 and p2 = p°2 × x2 x1 (mole fraction of CH3OH) = ((40/32)/(40)32+(60)46 = 0.49 [Mol. wt. of CH/3)OH = 32, Mol. wt. of C2H5OH = 46] x2 (Mole fraction of C2H5OH) = ((60/40)/(40)32+(60)46 = 0.51 p/1) = Partial vapour pressure of CH3OH = 88.7 × 0.49 = 43.46 mm Hg p2 = Partial vapour pressure of C2H5OH = 44.5 × 0.51 = 22.69 mm Hg PTotal =p1 +p2 = (43⋅46 + 22⋅69) = 66⋅15 mm Hg Mole fraction of CH3OH in vapour = (43.46/66.15) = 0.657

$43.46 mm$ and $0.51$

$66.15 mm$ and $0.657$

$66.15 mm$ and $0.791$

$70.59 mm$ and $0.657$

Q. The vapour pressures of ethanol and methanol are 44.5mmHg and 88.7mmHg respectively. An ideal solution is formed at the same temperature by mixing 60g of ethanol with 40g of methanol. The total vapour pressure of the solution and the mole fraction of methanol in the vapour are respectively

43.46mm and 0.51

66.15mm and 0.657

66.15mm and 0.791

70.59mm and 0.657

$43.46 mm$ and $0.51$

$66.15 mm$ and $0.657$

$66.15 mm$ and $0.791$

$70.59 mm$ and $0.657$