
What is the vapour pressure of the solution containing 34.2 g of sucrose per 100 g of water at 25$^0C$? The vapour pressure of water at 25$^0C$ is 23.75 mm.
a.) 20.3 mm
b.) 23.10 mm
c.) Unpredictable
d.) 23.33 mm
Answer
575.1k+ views
Hint: The vapour pressure of a solution may be defined as the product of partial pressure of a component multiplied by its mole fraction.
Mathematically, it can be written as -
P = ${P_0}{x_A}$
Where ‘P’ = pressure of the solution
${P_0}$= partial pressure of component
${x_A}$= mole fraction of component
First, we will find the mole fraction of each component and then vapour pressure.
Complete answer:
First, let us write what is given to us and what we need to find out.
Given :
Mass of sucrose = 34.2 g
Mass of water = 100 g
vapour pressure of water = 23.75 mm.
Temperature = 25$^0C$
To find :
Vapour pressure of the solution
We know that vapour pressure of a solution is a product of partial pressure of a component multiplied by its mole fraction.
Mathematically, it can be written as -
P = ${P_0}{x_A}$
Where ‘P’ = pressure of the solution
${P_0}$= partial pressure of component
${x_A}$= mole fraction of component
First, we will find the mole fraction of each component and then vapour pressure.
So, we have two components -
Let the water be component A and the sucrose be component B.
We can find the number of moles of each component.
We have, number of moles = $\dfrac{{Given{\text{ mass}}}}{{Molar{\text{ mass}}}}$
So, the number of moles of water, ${n_A} = \dfrac{{100}}{{18}}$
number of moles of water, ${n_A}$= 5.55 mole
number of moles of sucrose, ${n_B} = \dfrac{{34.2}}{{342}}$
number of moles of sucrose, ${n_B}$= 0.1 mole
Now, the mole fraction of water is as-
${x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$
${x_A} = \dfrac{{5.55}}{{5.55 + 0.1}}$
${x_A}$= 0.982
So, the vapour pressure is -
P = ${P_0}{x_A}$
P = 23.75$ \times $0.982
P = 23.33 mm
Thus, the correct answer is the option d.).
Note:
It must be noted that the mole fraction of all the components in a mixture is 1. Thus, the value of mole fraction of any component is from 0 to 1. The number of moles can be any number. It depends upon the amount of substance given and molar mass of substance.
Mathematically, it can be written as -
P = ${P_0}{x_A}$
Where ‘P’ = pressure of the solution
${P_0}$= partial pressure of component
${x_A}$= mole fraction of component
First, we will find the mole fraction of each component and then vapour pressure.
Complete answer:
First, let us write what is given to us and what we need to find out.
Given :
Mass of sucrose = 34.2 g
Mass of water = 100 g
vapour pressure of water = 23.75 mm.
Temperature = 25$^0C$
To find :
Vapour pressure of the solution
We know that vapour pressure of a solution is a product of partial pressure of a component multiplied by its mole fraction.
Mathematically, it can be written as -
P = ${P_0}{x_A}$
Where ‘P’ = pressure of the solution
${P_0}$= partial pressure of component
${x_A}$= mole fraction of component
First, we will find the mole fraction of each component and then vapour pressure.
So, we have two components -
Let the water be component A and the sucrose be component B.
We can find the number of moles of each component.
We have, number of moles = $\dfrac{{Given{\text{ mass}}}}{{Molar{\text{ mass}}}}$
So, the number of moles of water, ${n_A} = \dfrac{{100}}{{18}}$
number of moles of water, ${n_A}$= 5.55 mole
number of moles of sucrose, ${n_B} = \dfrac{{34.2}}{{342}}$
number of moles of sucrose, ${n_B}$= 0.1 mole
Now, the mole fraction of water is as-
${x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$
${x_A} = \dfrac{{5.55}}{{5.55 + 0.1}}$
${x_A}$= 0.982
So, the vapour pressure is -
P = ${P_0}{x_A}$
P = 23.75$ \times $0.982
P = 23.33 mm
Thus, the correct answer is the option d.).
Note:
It must be noted that the mole fraction of all the components in a mixture is 1. Thus, the value of mole fraction of any component is from 0 to 1. The number of moles can be any number. It depends upon the amount of substance given and molar mass of substance.
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