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InterviewSolution
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100 g of liquid A (molar mass "140g mol"^(-1)) was dissolved in 1000 g of liquid B (molar mass "180 g mol"^(-1)). The vapour pressure of pure liquid B was found o be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr. |
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Answer» Solution :`"No. of moles of liquid A (solute)"=("100 G")/("140 g mol"^(-1))=(5)/(7)"MOLE"` `"No. of moles of lliquid B (solvent)"=("1000 g")/("180 g mol"^(-1))=(50)/(9)" mole"` `THEREFORE"Mole fraction of A in the solution "(x_(A))=(5//7)/(5//7+50//9)=(5//7)/(395//63)=(5)/(7)xx(63)/(395)=(45)/(395)=0.114` `therefore"Mole fraction of B in the solution "(x_(B))=1-0.114=0.886` `"Also, given"p_(B)^(@)="500 torr"` `"Applying Raoult's law,"p_(A)=x_(A)p_(A)^(@)=0.114xxp_(A)^(@)"...(i)"` `p_(B)=x_(B)p_(B)^(@)=0.886xx500="443 torr"` `p_("Total")=p_(A)+p_(B)` `475=0.114p_(A)^(@)+443"or"p_(A)^(@)=(475-443)/(0.114)="280.7 torr"` Substituting this value in eqn. (i) we get `""p_(A)=0.114xx"280.7 torr = 32 torr."` |
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