Question

X, Y, and Z are three volatile liquids, which when mixed make an ideal solution. At a given temperature $\text{T}_0$, the constituents X, Y, and Z have pure vapour pressures in the ratio 3:2:1. A mixture prepared with mole-fractions of X, Y, and Z in the ratio 1:2:3 starts to boil at temperature $\text{T}_0$ at 1.5 atm pressure. At temperature $\text{T}_0$, the partial vapour pressure (in atm) of Y is:

• A. $\frac{3}{5}$
• B. $\frac{9}{10}$
• C. $\frac{1}{6}$
• D. $\frac{9}{20}$

Hint:

Double check your calculations by verifying that the sum of the partial pressures equals the total pressure:
\[ P_{\text{X}} = 0.45\ \text{atm}, \quad P_{\text{Y}} = 0.60\ \text{atm}, \quad P_{\text{Z}} = 0.45\ \text{atm} \implies \sum P = 1.50\ \text{atm}. \]

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to find the partial vapour pressure of liquid Y in an ideal three-component liquid mixture that starts to boil at a given total pressure of $1.5\ \text{atm}$.

Step 2: Key Formula or Approach:
1. According to Raoult's Law, the total vapour pressure ($P_{\text{total}}$) of an ideal solution containing components X, Y, and Z is:
\[ P_{\text{total}} = x_{\text{X}} P_{\text{X}}^0 + x_{\text{Y}} P_{\text{Y}}^0 + x_{\text{Z}} P_{\text{Z}}^0 \]
2. The partial vapour pressure of any component $i$ is:
\[ P_i = x_i P_i^0 \]

Step 3: Detailed Explanation:

• Let us define the pure vapour pressures of X, Y, and Z based on their given ratio of $3:2:1$:
\[ P_{\text{X}}^0 = 3k, \quad P_{\text{Y}}^0 = 2k, \quad P_{\text{Z}}^0 = 1k \]
where $k$ is a constant of proportionality.

• The mole fractions of the constituents in the mixture are in the ratio $1:2:3$. Therefore, we can write:
\[ x_{\text{X}} = \frac{1}{1+2+3} = \frac{1}{6} \]
\[ x_{\text{Y}} = \frac{2}{1+2+3} = \frac{2}{6} = \frac{1}{3} \]
\[ x_{\text{Z}} = \frac{3}{1+2+3} = \frac{3}{6} = \frac{1}{2} \]

• A liquid mixture starts to boil when its total vapour pressure equals the external pressure. Thus:
\[ P_{\text{total}} = 1.5\ \text{atm} \]

• Substituting these values into Raoult's Law:
\[ 1.5 = \left(\frac{1}{6} \times 3k\right) + \left(\frac{1}{3} \times 2k\right) + \left(\frac{1}{2} \times 1k\right) \]
\[ 1.5 = \frac{1}{2}k + \frac{2}{3}k + \frac{1}{2}k \]
\[ 1.5 = k + \frac{2}{3}k = \frac{5}{3}k \]

• Solving for the constant $k$:
\[ k = \frac{1.5 \times 3}{5} = \frac{4.5}{5} = 0.9\ \text{atm} \]

• Now, let us calculate the partial vapour pressure of Y:
\[ P_{\text{Y}} = x_{\text{Y}} P_{\text{Y}}^0 = \frac{1}{3} \times 2k = \frac{1}{3} \times 2(0.9) = 0.6\ \text{atm} \]

• Converting $0.6$ to a fraction:
\[ P_{\text{Y}} = 0.6 = \frac{6}{10} = \frac{3}{5}\ \text{atm} \]


Step 4: Final Answer:
Therefore, the partial vapour pressure of Y is $\frac{3}{5}\ \text{atm}$ (Option A).