Question

Calculate the vapor pressure of an aqueous solution of $1.0$ molal glucose solution as ${100^ \circ }C$

Answer 1

Hint: Use Raoult’s law for dilute solutions,as it is a relation between vapour pressure and mole fraction of solute. Modify the equation to relate relative lowering of vapour pressure and molality. Use this modified formula to solve the question.

Complete step by step answer:
Let us write the values which are given to us in the question and then try to identify the appropriate formula to be used.
$mass\,of\,solvent = 18g$, since the question mentions and aqueous solvent.
$molality = 1.0molal$
${P_ \circ } = 760mm\,at\,{100^ \circ }C$ , this is a standard value for vapor pressure of water at ${100^ \circ }C$ .
We need to find the vapor pressure of the solution after glucose has been added to the water.
This means there will be relative lowering in vapor pressure.
The relative lowering in Vapor pressure is given by Raoult's Law.
Raoult’s law of lowering of vapor pressure is equal to the mole fraction of the solute.
Hence the formula of Raoult’s law states that:
\[\dfrac{{{P_ \circ } - {P_S}}}{{{P_ \circ }}} = mole\,fraction\]
$Mole\,fraction\,of\,solute = \dfrac{{moles\,of\,solute}}{{moles\,of\,solute + \,moles\,of\,solvent}}$
Since this is a dilute solution, number of moles of solute in the denominator can be ignored and the new formula becomes:
$Mole\,fraction\,of\,solute = \dfrac{{Moles\,of\,solute}}{{Moles\,of\,solvent}}$
We can modify this equation as we know, $Moles\,of\,solvent = \dfrac{{Mass\,of\,solvent}}{{Molecular\,mass\,of\,solvent}}$ .

$\therefore mole\,fraction = \dfrac{{moles\,of\,solute}}{{moles\,of\,solvent}} \Rightarrow \dfrac{{moles\,of\,solute}}{{mass\,of\,solvent}} \times Molecular\,mass\,of\,solvent$

we know that; $molality = \dfrac{{number\,of\,moles\,of\,solute}}{{Mass\,of\,solvent}}$
By substituting this value in the above equation, we get:
$Mole\,fraction = Molality \times Molecular\,mass\,of\,solvent$
Substituting this value of mole fraction in the Raoult’s law, we get
\[\dfrac{{{P_ \circ } - {P_S}}}{{{P_ \circ }}} = molality \times mass\,of\,solvent\]
Now, Let us substitute the information given to us in the question:
Given:
$mass\,of\,solvent = mass\,of\,{H_2}O$
$mass\,of\,{H_2}O = Mass\,of\,H \times 2 + Mass\,of\,O$
$mass\,of\,{H_2}O = 1 \times 2 + 16 \Rightarrow 18g$
 ${P_ \circ } = 760mm\,at\,{100^ \circ }C$ and $mass\,of\,solvent = 18g$ $molality = 1.0molal$
By substituting these values in the above equation, determine the value for ${P_S}$
$\dfrac{{760mm - {P_S}}}{{760mm}} = 0.1 \times 18g$
Solving for the value of ${P_S}$ we get:
${P_S} = 746.32mm$

Note: The relative lowering of vapor pressure is a colligative property , which means it is dependent upon the amount of moles of the solution. Hence, It can be used for determination of the Van’t Hoff Factor. It is also an important tool for determining the molecular weight of Solutes in the solvent, this can help in identifying whether the solute has undergone any changes in the solution such as association and dissociation.