Question

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

Answer 1

The concept of Raoult's Lawand the role of it in lowering the vapour pressure of a non volatile solute is to be used in this question. Use the data given in the question and substitute it in the Raoult’s Law formula.

Complete step by step solution:
Let us see what Raoult’s law is. The French Chemist Francois Marie Raoult in 1886 did many of the experiments in order to study the vapour pressure of a number of binary solutions. He proposed a generalisation which is known as Raoult's law which states that, The mole fraction of the solvent is directly proportional to the vapour pressure of a solution containing non volatile solute on the basis of the results of the experiments. In case of solution containing two components (volatile solvent) and B (non-volatile solute) the vapour pressure of solution is given as:

& {{p}_{a}}\propto {{x}_{a}} \\\ & {{p}_{a}}=k{{x}_{a}} \\\ \end{aligned}$$ For solutions that obeys Raoult's law generally at all concentrations its vapour pressure would vary linearly from zero to the vapour pressure of pure solvent. For a solution that contains non-volatile solute, the relative lowering of vapour pressure is equal to mole fraction of the solute, at a given temperature. If mole fraction of solute is ${{x}_{B}}$, then we can write that ${{x}_{A}}+{{x}_{B}}=1\,or\,{{x}_{A}}=1-{{x}_{B}}$, and also,${{P}_{A}}={{p}_{A}}^{0}{{x}_{A}}$. Thus by combining the two given equations, we can write: $$\begin{aligned} & {{p}_{A}}={{p}_{A}}^{0}(1-{{x}_{B}})={{p}_{A}}^{0}-{{p}_{A}}^{0}{{x}_{B}} \\\ & or,\,{{p}_{A}}^{0}-{{p}_{A}}={{p}_{A}}^{0}{{x}_{B}} \\\ & or,\,\dfrac{{{p}_{A}}^{0}-{{p}_{A}}}{{{P}_{A}}^{0}}={{x}_{B}} \\\ \end{aligned}$$ The numerator is the lowering of the vapour pressure, and the ratio is called relative lowering of vapour pressure. This is Raoult's Law. As, ${{x}_{B}}$is mole fraction, so we can write: $$\begin{aligned} & \dfrac{100-80}{100}=\dfrac{{{w}_{2}}/40}{114/114+{{w}_{2}}/40}\,\,\,\,\,\,\,molar\,mas{{s}_{{{C}_{8}}{{H}_{18}}}}=114g\,mo{{l}^{-1}} \\\ & So,\,\dfrac{20}{100}=\dfrac{{{w}_{2}}/40}{1+{{w}_{2}}/40}=\dfrac{1}{5}(1+\dfrac{{{w}_{2}}}{40}) \\\ \end{aligned}$$ $$\begin{aligned} & 40+{{w}_{2}}=5{{w}_{2}} \\\ & \Rightarrow 40=4{{w}_{2}} \\\ & \Rightarrow {{w}_{2}}=10g \\\ \end{aligned}$$ **So, we obtain the mass of the non volatile solute as 10 grams which is our required answer.** **NOTE:** If in case a situation occurs that the non volatile components of Raoult’s Law are applied not to liquids, but in gases, then the formula becomes similar to Henry’s Law. Hence, Raoult’s law can be said as a special case of Henry’s Law.${{p}_{A}}={{p}_{A}}^{0}{{x}_{a}}/{{p}_{A}}={{K}_{H}}x$.