In a solvent S, a compound B is partially dissociated into C and D as given below:

B ⇌ 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 _______.

The correct answer is 33.3.

Let us solve the problem step-by-step to determine the mole percent of B that has been dissociated.

1. Calculation of the ebullioscopic constant (K_b) of solvent S:
The relation for the ebullioscopic constant (K_b) is given by:

$K_b=\frac{R\cdot {\left(T_b\right)}^2\cdot M_S}{1000\cdot mo><mi>\Delta </mi><msub><mi>H</mi><mrow><mi>v</mi><mi>a</mi><mi>p</mi></mrow></msub>}$

Given:
Standard boiling point of solvent S, T_b = 400 K
Enthalpy of vaporization of solvent S, ΔH_vap = 10R J mol⁻¹
Molar mass of solvent S = M_S

Substituting these values into the expression for K_b:

$K_b=\frac{R\cdot {\left(400\right)}^2\cdot M_S}{1000\cdot 10R}=\frac{160000\cdot M_S}{10000}=16\cdot M_S$

2. Elevation in boiling point (ΔT_b):
ΔT_b = T_solution - T_b = 408 K - 400 K = 8 K

3. Molality of solute B (m):
Given that the initial concentration of B in solvent S is 0.25% (mass/mass):
Mass of B = 0.25 g
Mass of solvent S = 100 g - 0.25 g ≈ 100 g = 0.1 kg
Molar mass of B, M_B = 10 × M_S

The molality m is given by:

$m=\frac{\text{Moles of B}}{\text{Mass of solvent in kg}}=\frac{0.25/M_B}{0.1}=\frac{2.5}{10\cdot M_S}=\frac{0.25}{M_S}$

4. Determination of van 't Hoff factor (i):
Using the formula for elevation in boiling point:

$\Delta T_b=i\cdot K_b\cdot m$

Substitute the values into the equation:

$8=i\cdot (16\cdot M_S)\cdot \left(\frac{0.25}{M_S}\right)$

$8=i\cdot 4\Rightarrow i=2$

5. Calculation of degree of dissociation (α) and mole percent:
For the reaction:
B ⇌ 2C + 2D
Initial moles: 1 mole of B, 0 of C, 0 of D.
At equilibrium: (1 - α) moles of B, 2α moles of C, 2α moles of D.
Total moles at equilibrium = (1 - α) + 2α + 2α = 1 + 3α.

Therefore, the van 't Hoff factor is:

$i=1+3\alpha$

Equating the value of i:

$1+3\alpha =2\Rightarrow 3\alpha =1\Rightarrow \alpha =\frac{1}{3}\approx 0.3333$

The mole percent of B that has dissociated is:

$\text{Mole percent}=\alpha \cdot 100\%=\frac{1}{3}\cdot 100\%=33.3\%$