
At 25$^{ 0 }$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 25$^{ 0 }$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
Answer
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Hint: 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,
$P_{ t }=P_{ A }^{ 0 }X_{ A }+P_{ B }^{ 0 }X_{ B }$
Where,
$P_{ t }$ = total vapour pressure
$P_{ A }^{ 0 }$ = vapour pressure pure benzene
$X_{ A }$ = mole fraction of benzene
$P_{ B }^{ 0 }$ = vapour pressure of pure ethyl alcohol
$X_{ B }$ = 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 = $\dfrac { weight\quad of\quad benzene }{ Molar\quad mass\quad of\quad benzene } =\dfrac { 10 }{ 78 }$
Moles of benzene = 0.128 moles
Moles of ethyl alcohol = $\dfrac { weight\quad of\quad ethyl alcohol }{ Molar\quad mass\quad of\quad ethyl alcohol } =\dfrac { 10 }{ 46 }$
Moles of ethyl alcohol = 0.217
Now, from these values, we need to calculate mole fraction,
Mole fraction of benzene ($X_{ A }$) = $\dfrac { moles\quad of\quad A\quad }{ Total\quad moles } =\dfrac { 0.128 }{ 0.345 }$ = 0.371
Mole fraction of ethyl alcohol ($X_{ B }$) = 1-$X_{ A }$ = 0.629
Now, put these values in the formula
$P_{ t }=P_{ A }^{ 0 }X_{ A }+P_{ B }^{ 0 }X_{ B }$
$P_{ t }$ = 100x0.371 + 44x0.629
$P_{ t }$ = 37.1 + 27.675
$P_{ t }$ = 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.
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,
$P_{ t }=P_{ A }^{ 0 }X_{ A }+P_{ B }^{ 0 }X_{ B }$
Where,
$P_{ t }$ = total vapour pressure
$P_{ A }^{ 0 }$ = vapour pressure pure benzene
$X_{ A }$ = mole fraction of benzene
$P_{ B }^{ 0 }$ = vapour pressure of pure ethyl alcohol
$X_{ B }$ = 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 = $\dfrac { weight\quad of\quad benzene }{ Molar\quad mass\quad of\quad benzene } =\dfrac { 10 }{ 78 }$
Moles of benzene = 0.128 moles
Moles of ethyl alcohol = $\dfrac { weight\quad of\quad ethyl alcohol }{ Molar\quad mass\quad of\quad ethyl alcohol } =\dfrac { 10 }{ 46 }$
Moles of ethyl alcohol = 0.217
Now, from these values, we need to calculate mole fraction,
Mole fraction of benzene ($X_{ A }$) = $\dfrac { moles\quad of\quad A\quad }{ Total\quad moles } =\dfrac { 0.128 }{ 0.345 }$ = 0.371
Mole fraction of ethyl alcohol ($X_{ B }$) = 1-$X_{ A }$ = 0.629
Now, put these values in the formula
$P_{ t }=P_{ A }^{ 0 }X_{ A }+P_{ B }^{ 0 }X_{ B }$
$P_{ t }$ = 100x0.371 + 44x0.629
$P_{ t }$ = 37.1 + 27.675
$P_{ t }$ = 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.
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