Question

In an ideal binary solution of two liquids a and b the mole fraction of the gases in vapour phase were found to be equal

Answer 1

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▪Question:-

The vapour pressures of ethanol and methanol are 44.5 mm 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. Calculate the total vapour pressure for solution and the mole fraction of methanol in the vapour.

Solution:
Mol. mass of ethyl alcohol = C2H2OH = 46

No. of moles of ethyl alcohol = 60/46 = 1.304

Mol. mass of methyl alcohol = CH3OH = 32

No. of moles of methyl alcohol = 40/32 = 1.25

‘XA’, mole fraction of ethyl alcohol = 1.304/1.304+1.25 = 0.5107

‘XB’, mole fraction of methyl alcohol = 1.25/1.304+1.25

><Partial pressure of ethyl alcohol = XA. pA0 = 0.5107 × 44.5 = 22.73 mm Hg

><Partial pressure of methyl alcohol = XB.pA0 =0.4893 × 88.7 = 43.73 m Hg

><Total vapour pressure of solution = 22.73 + 43.40 = 66.13 mm Hg

▪Mole fraction of methyl alcohol in the vapour = Partial pressure of CH3OH/Total vapour pressure = 43.40/66.13 = 0.6563

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