Question

How would you determine the vapour pressure of a solution at $25{}^\circ C$ that contain $76.6g$ of glucose $({{C}_{6}}{{H}_{12}}{{O}_{6}})$ in $250.0mL$ of water? The vapour pressure of pure water at $25{}^\circ C$ is $23.8torr$.

Answer 1

The vapour pressure of a substance in a solution is the pressure exerted by that substance in the solution.
The vapour pressure of a solution can be calculated by using the mole fraction of the solvent along with the vapour pressure of the non-volatile solvent.

Complete step-by-step answer: In the given question some values of the quantities are given to us, and so we will write down all those quantities and see how we could use them to determine the vapour pressure of the solution.
The temperature is given as $25{}^\circ C$, and the amount of solute which is glucose is given as $76.6g$ and the amount of water in terms of millilitre is given as $250.0mL$. The vapour pressure of pure water is also provided to us in terms of torr which is $23.8torr$.
Now, we can use the volume and density of water in order to get the mass of water. As we know that the density is mass of a substance present per unit volume of the substance. And also we know that the density of water at $25{}^\circ C$ is, $\rho =0.99705{ }gm{{L}^{-1}}$. So, now we can calculate the mass of water as,
$250.0mL\times \dfrac{0.99705g}{1mL}=249.26g$
We can see that the mass of water came out to be $249.26g$. Now, as we know that the water is a non-volatile compound so the vapour pressure of the considered solution would depend only on the value of the mole fraction of that sample water and the vapour pressure of the distilled water.
So, now we will calculate the number of moles of the glucose and the water individually, by using the mass and the molar mass of both. We know that the number of moles of a substance is the mass of that substance which is present per molar mass of it.
Now, the number of moles of glucose are,
$76.6g\times \dfrac{1{ }mole{ }glucose}{180.156g}=0.4252moles$
Where, the molar mass of the glucose is $180.156gmo{{l}^{-1}}$ and the mass of glucose is given as $76.6g$.
Similarly, the number of moles of water is,
$249.26g\times \dfrac{1{ }mole{ }water}{18.015g}=13.836moles$
Where, the molar mass of water is $18.015gmo{{l}^{-1}}$ and the mass of water is $249.26g$.
So, the mole fraction of the given water would be the ratio of the number of moles of water to that of the total number of moles which is present in the solution. So, the total number of moles in the solution is,
${{n}_{total}}={{n}_{glucose}}+{{n}_{water}}$
Where, ‘n’ represents the number of moles and the subscripts indicate the specific substances. Now, we will put the calculated values of the number of moles to find out the total number of moles. We get,
${{n}_{total}}=0.4252+13.836=14.261{ }moles$
Now, we know the total number of moles we will simply calculate the mole fraction of water, we get,
${{\chi }_{water}}=\dfrac{{{n}_{water}}}{{{n}_{glucose}}+{{n}_{water}}}$
Now, we will put the values of the quantities, in order to get the mole fraction, we get,
${{\chi }_{water}}=\dfrac{13.836moles}{14.261moles}=0.9702$
So, now the vapour pressure of the considered solution can be calculated using this value of mole fraction along with the given value of vapour pressure of the pure water, we get,
${{P}_{sol}}={{\chi }_{water}}\times {{P}^{\circ }}_{water}$
Here the value of vapour pressure of solution is unknown, and rest of the values are known, so we will substitute those known values and we get,
${{P}_{sol}}=0.9702\times 23.8torr=23.091torr$
So, the vapour pressure of the solution becomes $23.1torr$ approximately.

Note: The vapour pressure of the solution is equal to the multiplication of the mole fraction of the non-volatile component of the solution and the vapour pressure of the water.
Mole fraction of a substance is the number of moles of that substance present in the total number of moles present in the solution.