JEE Challenger
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ChemHardAdv 2026 • Paper 2

Mole Percent Dissociation of Compound B in Solvent S

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

B⇌2C+2D\mathbf{B} \rightleftharpoons 2\mathbf{C} + 2\mathbf{D}

B\mathbf{B}, C\mathbf{C} and D\mathbf{D} are non-volatile in nature. The molar mass of B\mathbf{B} is 1010 times the molar mass of S\mathbf{S}. The standard boiling point and the standard enthalpy of vaporization of S\mathbf{S} are 400 K400\text{ K} and 10R J mol−110R\text{ J mol}^{-1}, respectively (RR is the gas constant in J K−1 mol−1\text{J K}^{-1}\text{ mol}^{-1}). A solution of B\mathbf{B} in S\mathbf{S} with an initial concentration of B\mathbf{B} as 0.25%0.25\% (mass/mass) has a boiling point of 408 K408\text{ K} at 1 bar1\text{ bar} pressure. In this solution, the mole percent of B\mathbf{B} that has been dissociated is _____.

Official Numerical Answer31.6 to 33.33

Step-by-Step Solution

To find the mole percent of compound B\mathbf{B} that has dissociated in solvent S\mathbf{S}, we proceed step-by-step:

1. Dissociation Reaction and Van 't Hoff Factor (ii)

The dissociation equilibrium of compound B\mathbf{B} is given by: B⇌2C+2D\mathbf{B} \rightleftharpoons 2\mathbf{C} + 2\mathbf{D}

Let α\alpha be the degree of dissociation of B\mathbf{B}.

Initial moles:n000Equilibrium moles:n0(1−α)2n0α2n0α\begin{array}{lcccc} \text{Initial moles:} & n_0 & 0 & 0 \\ \text{Equilibrium moles:} & n_0(1-\alpha) & 2n_0\alpha & 2n_0\alpha \end{array}

The total moles of solute species at equilibrium is: ntotal solute=n0(1−α)+2n0α+2n0α=n0(1+3α)n_{\text{total solute}} = n_0(1-\alpha) + 2n_0\alpha + 2n_0\alpha = n_0(1 + 3\alpha)

Thus, the van 't Hoff factor ii is: i=ntotal soluten0=1+3αi = \frac{n_{\text{total solute}}}{n_0} = 1 + 3\alpha

2. Ebullioscopic Constant (KbK_b) of Solvent S\mathbf{S}

The molal elevation constant KbK_b for a solvent is given by: Kb=R(Tb∘)2MS1000⋅ΔvapHK_b = \frac{R (T_b^\circ)^2 M_{\mathbf{S}}}{1000 \cdot \Delta_{\text{vap}} H}

Given:

  • Standard boiling point of solvent, Tb∘=400 KT_b^\circ = 400\text{ K}
  • Standard enthalpy of vaporization, ΔvapH=10R J mol−1\Delta_{\text{vap}} H = 10 R\text{ J mol}^{-1}
  • Molar mass of solvent = MSM_{\mathbf{S}}

Substituting these values into the formula for KbK_b: Kb=R⋅(400)2⋅MS1000⋅10R=160000⋅MS10000=16MS K kg mol−1K_b = \frac{R \cdot (400)^2 \cdot M_{\mathbf{S}}}{1000 \cdot 10 R} = \frac{160000 \cdot M_{\mathbf{S}}}{10000} = 16 M_{\mathbf{S}} \text{ K kg mol}^{-1}

3. Molality (mm) of the Solution

In a 0.25%0.25\% (mass/mass) solution of B\mathbf{B} in S\mathbf{S}:

  • Mass of solute B=0.25 g\mathbf{B} = 0.25\text{ g}
  • Mass of solvent S=100 g−0.25 g=99.75 g=0.09975 kg\mathbf{S} = 100\text{ g} - 0.25\text{ g} = 99.75\text{ g} = 0.09975\text{ kg}

Given that the molar mass of B\mathbf{B} is MB=10MSM_{\mathbf{B}} = 10 M_{\mathbf{S}}: Moles of B,nB=0.25MB=0.2510MS=0.025MS\text{Moles of } \mathbf{B}, n_{\mathbf{B}} = \frac{0.25}{M_{\mathbf{B}}} = \frac{0.25}{10 M_{\mathbf{S}}} = \frac{0.025}{M_{\mathbf{S}}}

The molality mm of the solution is: m=nBMass of solvent (in kg)=0.025/MS0.09975=13.99MSm = \frac{n_{\mathbf{B}}}{\text{Mass of solvent (in kg)}} = \frac{0.025 / M_{\mathbf{S}}}{0.09975} = \frac{1}{3.99 M_{\mathbf{S}}}

4. Elevation in Boiling Point and Degree of Dissociation (α\alpha)

The elevation in boiling point ΔTb\Delta T_b is: ΔTb=Tb−Tb∘=408 K−400 K=8 K\Delta T_b = T_b - T_b^\circ = 408\text{ K} - 400\text{ K} = 8\text{ K}

Using the colligative property relation ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot m: 8=i⋅(16MS)⋅(13.99MS)8 = i \cdot \left(16 M_{\mathbf{S}}\right) \cdot \left(\frac{1}{3.99 M_{\mathbf{S}}}\right)

8=i⋅163.998 = i \cdot \frac{16}{3.99}

i=8×3.9916=3.992=1.995i = \frac{8 \times 3.99}{16} = \frac{3.99}{2} = 1.995

Equating ii to 1+3α1 + 3\alpha: 1+3α=1.9951 + 3\alpha = 1.995 3α=0.995  ⟹  α=0.9953≈0.331673\alpha = 0.995 \implies \alpha = \frac{0.995}{3} \approx 0.33167

The mole percent of B\mathbf{B} that has dissociated is: Mole percent=α×100%≈33.17%\text{Mole percent} = \alpha \times 100\% \approx 33.17\%

(Note: Using the dilute solution approximation where mass of solvent ≈100 g\approx 100\text{ g} gives i=2  ⟹  α=13≈33.33%i = 2 \implies \alpha = \frac{1}{3} \approx 33.33\%)

Final Answer: The mole percent of B\mathbf{B} that has been dissociated is 33.17 (or 33.33 using approximation).