Question

Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.
Given: The gas constant \(R = 0.08 \,\text{L atm K}^{-1}\text{ mol}^{-1}\)
Molar mass of A is \(50 \,\text{g mol}^{-1}\)
Molar mass of B is \(57 \,\text{g mol}^{-1}\)
Density of liquid B at \(300\,\text{K}\) is \(0.5 \,\text{g mL}^{-1}\)
\(1\,\text{atm} = 760\,\text{mm Hg}\)

At 300 K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is _______.

Hint:

Use density to find molar volume in liquid phase and ideal gas equation to find molar volume in vapour phase.

Answer 1

Correct Answer: 210.5

Step 1: Understanding the Question:
We need to calculate the ratio of molar volume of pure liquid \(B\) in vapour phase to its molar volume in liquid phase at \(300\,K\).
Given: \[ R=0.08\ \text{L atm K}^{-1}\text{mol}^{-1} \] \[ M_B=57\ g\ mol^{-1} \] \[ \rho_B=0.5\ g\ mL^{-1} \]

Step 2: Key Formula or Approach:
For vapour phase: \[ V_{vap}=\frac{RT}{P} \] For liquid phase: \[ V_{liq}=\frac{\text{Molar mass}}{\text{Density}} \] Required ratio: \[ \frac{V_{vap}}{V_{liq}} \]

Step 3: Detailed Explanation:
(i) Molar volume in liquid phase \[ V_{liq}=\frac{57}{0.5} \] \[ =114\ mL \] \[ =0.114\ L \] (ii) Molar volume in vapour phase For ideal gas: \[ V_{vap}=\frac{RT}{P} \] At: \[ T=300\,K,\qquad P=1\ atm \] \[ V_{vap}=\frac{0.08\times300}{1} \] \[ =24\ L \] (iii) Ratio \[ \frac{V_{vap}}{V_{liq}} = \frac{24}{0.114} \] \[ \approx210.5 \]

Step 4: Final Answer:
The required ratio is: \[ \boxed{210.5} \]