The ideal solutions formed by mixing two liquids $\mathrm{A}$ and $\mathrm{B}$ at $300 \mathrm{~K}$ in the…
The ideal solutions formed by mixing two liquids $\mathrm{A}$ and $\mathrm{B}$ at $300 \mathrm{~K}$ in the molar ratio of $1: 1$ and $1: 2$ have vapour pressures of $400 \mathrm{~mm}$ and $350 \mathrm{~mm}$ respectively. At the same temperature, the vapour pressures of pure liquids $\mathrm{A}$ and $\mathrm{B}$ in $\mathrm{mm}$ respectively are:
$250,550$
$500,500$
$550,250$
$350,450$
Solution
Step 1: Set up equations based on Raoult's law. Raoult's law for a solution of two volatile liquids A and B is given by $P_{\text{total}}=x_{A}P_{A}^{0}+x_{B}P_{B}^{0}$, where $P_{\text{total}}$ is the total vapor pressure, $x_{A}$ and $x_{B}$ are the mole fractions, and $P_{A}^{0}$ and $P_{B}^{0}$ are the vapor pressures of the pure liquids. For the molar ratio of 1:1, the mole fractions are $x_{A}=\frac{1}{1+1}=0.5$ and $x_{B}=\frac{1}{1+1}=0.5$. The vapor pressure is $400\ \text{mm}$. This gives us the equation: $400=0.5P_{A}^{0}+0.5P_{B}^{0}$. This simplifies to: $800=P_{A}^{0}+P_{B}^{0}\quad (\text{Equation\ 1})$. For the molar ratio of 1:2, the mole fractions are $x_{A}=$\frac{1}{1+2}$=\frac{1}{3}$ and $x_{B}=\frac{2}{1+2}=\frac{2}{3}$. The vapor pressure is $350\ \text{mm}$. This gives us the equation: $350=\frac{1}{3}P_{A}^{0}+\frac{2}{3}P_{B}^{0}$. This simplifies to: $1050=P_{A}^{0}+2P_{B}^{0}\quad (\text{Equation\ 2})$.
Step 2: Solve the system of equations. We have two equations with two unknowns ($P_{A}^{0}$ and $P_{B}^{0}$): $P_{A}^{0}+P_{B}^{0}=800$ and $P_{A}^{0}+2P_{B}^{0}=1050$. Subtract Equation 1 from Equation 2: $(P_{A}^{0}+2P_{B}^{0})-(P_{A}^{0}+P_{B}^{0})=1050-800$, which gives $P_{B}^{0}=250$. Substitute the value of $P_{B}^{0}$ back into Equation 1 to find $P_{A}^{0}$: $P_{A}^{0}+250=800$, which simplifies to $P_{A}^{0}=800-250$ and $P_{A}^{0}=550$. The vapor pressures of pure liquids A and B are 550 mm and 250 mm, respectively.