Question
Question: At 25\(^{ 0 }\)C, the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol ...
At 250C, the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol is 44 torr. Assuming ideal behaviour, calculate the vapour pressure at 250C of a solution which contains 10g of each substance.
(A.) 33.775 torr
(B.) 54.775 torr
(C.) 64.775 torr
(D.) 60.775 torr
Solution
Try to recall the formula for the vapour pressure of a solution. It is the sum of the vapour pressure of pure solute multiplied by their mole fraction and the vapour pressure of pure solvent multiplied by their mole fraction.
_Complete step by step solution: _
We should know that vapour pressure is defined as the pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system.
As we know total vapour pressure,
Pt=PA0XA+PB0XB
Where,
Pt = total vapour pressure
PA0 = vapour pressure pure benzene
XA = mole fraction of benzene
PB0 = vapour pressure of pure ethyl alcohol
XB = mole fraction of ethyl alcohol
Here in this question, we have 10g of each ethyl alcohol and benzene.
First, we need to calculate their moles:
Moles of benzene = Molarmassofbenzeneweightofbenzene=7810
Moles of benzene = 0.128 moles
Moles of ethyl alcohol = Molarmassofethylalcoholweightofethylalcohol=4610
Moles of ethyl alcohol = 0.217
Now, from these values, we need to calculate mole fraction,
Mole fraction of benzene (XA) = TotalmolesmolesofA=0.3450.128 = 0.371
Mole fraction of ethyl alcohol (XB) = 1-XA = 0.629
Now, put these values in the formula
Pt=PA0XA+PB0XB
Pt = 100x0.371 + 44x0.629
Pt = 37.1 + 27.675
Pt = 64.775 torr
Therefore, we can conclude that the correct answer to this question is option C.
Note: We should know that the equilibrium vapour pressure or vapour pressure is an indication of a liquid's evaporation rate.