Comprehension:
Solutions
Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase. Let the total vapour pressure at this stage be ptotal and p1 and p2 be the partial vapour pressures of the two components 1 and 2 respectively. These partial pressures are related to the mole fractions x1 and x2 of the two components 1 and 2 respectively. The French chemist, Francois Marie Raoult (1886) gave the quantitative relationship between them. The relationship is known as the Raoult's law which states that for a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction (x1) present in solution.
On mixing, benzene and toluene form an ideal solution. At 88 °C, the vapour pressure of the two liquid components (benzene and toluene) are 957 mm Hg and 379.5 mm Hg respectively. The vapour pressure of the solution obtained by mixing 19.5 g of benzene and 46 g of toluene will be approximately: (molar mass of benzene = 78 g mol-1 and toluene = 92 g mol-1)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 572 mm Hg
Benzene and toluene form an ideal solution, so both components obey Raoult’s law: the partial vapour pressure of each equals its pure vapour pressure times its mole fraction in the liquid, and the total vapour pressure is the sum of the two:
\( \displaystyle p_{total}=p_B^{\circ}\,x_B+p_T^{\circ}\,x_T \)
So the masses must first be converted into moles and then into mole fractions.
\( \displaystyle n(\ce{C6H6})=\frac{19.5}{78}=0.25\ \mathrm{mol} \)
\( \displaystyle n(\ce{C6H5CH3})=\frac{46}{92}=0.50\ \mathrm{mol} \)
\( \displaystyle x_B=\frac{0.25}{0.25+0.50}=\frac{1}{3} \quad\quad x_T=\frac{2}{3} \)
\( \displaystyle p_{total}=957\times\frac{1}{3}+379.5\times\frac{2}{3} \)
\( \displaystyle p_{total}=319+253=572\ \mathrm{mm\ Hg} \)
So the vapour pressure of the solution is about 572 mm Hg.
Shortcut: There are twice as many moles of toluene as of benzene, so the answer is a weighted average: (957 + 2 × 379.5)/3 = 1716/3 = 572 mm Hg.