Question

State Raoult's law. Calculate the mass of non-volatile solute (Molar mass $\,40g/mo{l^{ - 1}}\,$) which should be dissolved in $\,114g\,$ of octane to reduce its vapour pressure to $\,80\% \,$.

Answer 1

The molar mass of octane is $\,114.2g/mol\,$. Number of moles can be calculated with the help of molar mass and then the mass of the non- volatile solute can be calculated using the appropriate formula of the relative lowering of vapor pressure which is based on the number of moles.

Complete step by step answer:
Let us first understand Raoult’s law;
Raoult’s law states that; The partial vapour pressure of any volatile substance of a solution is the product of vapour pressure of that pure substance and the mole fraction of the component in the solution.
${P_{solution\,}} = {P^0}X$
Where $\,P\,$ is the vapour pressure of the substance in solution, $\,{P^0}\,$ is the vapour pressure of pure substance and $\,X\,$ is the mole fraction of that substance in the solution.
As we know, the molar mass of octane is $\,114.2g/mol\,$.
$Number\,of\,moles = \dfrac{{Weight\,in\,grams}}{{Molar\,mass}}$
So, the number of moles can be calculated by using the above formula;
Number of moles of octane $\, = \dfrac{{114g}}{{114.2g/mol}}\,$
$\simeq 1mol\,$
Let us assume that we require $\,W\,grams$ of non- volatile solute. It is given that the molar mass of the non-volatile solute is $\,40g/mol\,$.
Therefore, the number of moles of non-volatile solute $\, = \dfrac{W}{{40}}\,$
Then, the mole fraction of non-volatile solute is;
$Mole{\text{ }}fraction(X) = \dfrac{{Moles\,of\,solute}}{{Total\,moles\,of\,solution}}$
So, $\,X = \dfrac{{\dfrac{W}{{40}}}}{{1 + \dfrac{W}{{40}}}}$
Now, the mass can be calculated from the relative lowering of vapor pressure which is related to the number of moles.
$\dfrac{{{P^0} - P}}{{{P^0}}} = X\,$
$\Rightarrow \dfrac{{100 - 80}}{{100}} = \dfrac{{\dfrac{W}{{40}}}}{{1 + \dfrac{W}{{40}}}}$
So, by subtracting and dividing the left-hand side and cross multiplying the denominator we get;
$0.2 = \dfrac{{\dfrac{W}{{40}}}}{{\dfrac{{40 + W}}{{40}}}}\,$
$0.2 = \dfrac{W}{{40 + W}}\,$
Then, let’s bring the denominator to the left side in-order to calculate the variable $\,W\,$
$40 + W = \dfrac{1}{{0.2}}W\,$
$40 + W = 5W\,$
Here we are in the last step now, which involves bringing the variable part to one side of the equation;
$40 = 4W\,$
$\therefore W = 10g\,$

So, $\,10\,grams\,$ of non-volatile solute is required.

Note: At equilibrium, the rate of evaporation of the solid or liquid is equal to the rate of condensation of the gas back to its original state. Both solids and liquids have a vapour pressure and, regardless of how much of the material is present, this pressure is unchanged.