Question

In a solvent S, a compound B is partially dissociated into C and D as given below:
B \(\rightleftharpoons\) 2C + 2D
B, C and D are non-volatile in nature. The molar mass of B is 10 times the molar mass of S. The standard boiling point and the standard enthalpy of vaporization of S are 400 K and 10R J mol\(^{-1}\), respectively (R is the gas constant in J K\(^{-1}\) mol\(^{-1}\)). A solution of B in S with an initial concentration of B as 0.25% (mass/mass) has a boiling point of 408 K at 1 bar pressure. In this solution, the mole percent of B that has been dissociated is _______.

Hint:

For dissociation reactions, first determine the van’t Hoff factor using the total number of particles formed after dissociation, then use colligative property equations.

Answer 1

Correct Answer: 33.3

Step 1: Understanding the Question:
Compound \(B\) partially dissociates as: \[ B \rightleftharpoons 2C+2D \] We are given:
• Standard boiling point of solvent \(S = 400\,K\)
• Enthalpy of vaporization: \[ \Delta H_{vap}=10R\ \text{J mol}^{-1} \]
• Initial concentration of \(B =0.25\%\) (mass/mass)
• Boiling point of solution: \[ 408\,K \] We need to calculate the percentage dissociation of \(B\).

Step 2: Key Formula or Approach:
Elevation in boiling point: \[ \Delta T_b=iK_bm \] where:
• \(i\) = van’t Hoff factor
• \(K_b\) = ebullioscopic constant
• \(m\) = molality Also, \[ K_b=\frac{RT_b^2M}{\Delta H_{vap}} \] For dissociation: \[ B \rightleftharpoons 2C+2D \] If degree of dissociation is \(\alpha\), then: \[ i=1+3\alpha \]

Step 3: Detailed Explanation:
(i) Calculate \(K_b\): \[ K_b=\frac{R(400)^2M}{10R} \] \[ =\frac{160000M}{10} \] \[ K_b=16000M \] where \(M\) is molar mass of solvent \(S\).
(ii) Calculate molality: Assume \(100\,g\) of solution. Mass of \(B\): \[ 0.25\,g \] Mass of solvent: \[ 99.75\,g=0.09975\,kg \] Given: \[ \text{Molar mass of }B=10M \] Moles of \(B\): \[ \frac{0.25}{10M} \] Thus, \[ m=\frac{0.25/(10M)}{0.09975} \] \[ m\approx \frac{0.2506}{M} \] (iii) Apply boiling point elevation equation: \[ \Delta T_b=408-400=8\,K \] \[ 8=i(16000M)\left(\frac{0.2506}{M}\right) \] \[ 8=i(4009.6) \] \[ i\approx2 \] (iv) Calculate degree of dissociation: Since: \[ i=1+3\alpha \] \[ 2=1+3\alpha \] \[ 3\alpha=1 \] \[ \alpha=\frac13 \] Percentage dissociation: \[ \frac13\times100 \] \[ =33.3\% \]

Step 4: Final Answer:
The mole percent of \(B\) dissociated is: \[ \boxed{33.3\%} \]