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Question: At 25\(^{ 0 }\)C, the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol ...

At 250^{ 0 }C, 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 250^{ 0 }C 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

Explanation

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+PB0XBP_{ t }=P_{ A }^{ 0 }X_{ A }+P_{ B }^{ 0 }X_{ B }
Where,
PtP_{ t } = total vapour pressure
PA0P_{ A }^{ 0 } = vapour pressure pure benzene
XAX_{ A } = mole fraction of benzene
PB0P_{ B }^{ 0 } = vapour pressure of pure ethyl alcohol
XBX_{ B } = 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 = weightofbenzeneMolarmassofbenzene=1078\dfrac { weight\quad of\quad benzene }{ Molar\quad mass\quad of\quad benzene } =\dfrac { 10 }{ 78 }
Moles of benzene = 0.128 moles
Moles of ethyl alcohol = weightofethylalcoholMolarmassofethylalcohol=1046\dfrac { weight\quad of\quad ethyl alcohol }{ Molar\quad mass\quad of\quad ethyl alcohol } =\dfrac { 10 }{ 46 }
Moles of ethyl alcohol = 0.217

Now, from these values, we need to calculate mole fraction,
Mole fraction of benzene (XAX_{ A }) = molesofATotalmoles=0.1280.345\dfrac { moles\quad of\quad A\quad }{ Total\quad moles } =\dfrac { 0.128 }{ 0.345 } = 0.371
Mole fraction of ethyl alcohol (XBX_{ B }) = 1-XAX_{ A } = 0.629

Now, put these values in the formula
Pt=PA0XA+PB0XBP_{ t }=P_{ A }^{ 0 }X_{ A }+P_{ B }^{ 0 }X_{ B }
PtP_{ t } = 100x0.371 + 44x0.629
PtP_{ t } = 37.1 + 27.675
PtP_{ t } ​= 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.