In a solvent S, a compound B is partially dissociated into C and D as given below : B 2C + 2D. B

Q: Why did we take the mass of the solvent as 99.75 g instead of 100 g?

Because the concentration is given as 0.25% by mass. This means 100 g of the **entire solution** contains 0.25 g of the solute. Therefore, the mass of the solvent alone is $100 - 0.25 = 99.75 \text{ g}$.

Q: How did we get $n=4$ for the van't Hoff factor?

Look at the balanced chemical equation: $B \rightleftharpoons 2C + 2D$. One mole of reactant B produces 2 moles of C and 2 moles of D. So, the total number of product particles from one reactant particle is $2 + 2 = 4$.

Q: What is the formula used for calculating $K_b$?

We used the thermodynamic relation $K_b = \frac{R (T_b^\circ)^2 M_S}{1000 \Delta H_{vap}}$. This formula relates the ebullioscopic constant to the universal gas constant, the boiling point of the pure solvent, its molar mass, and its enthalpy of vaporization.

Q: Why did the molar masses $M_B$ and $M_S$ cancel out?

The problem cleverly states that the molar mass of B is 10 times that of S ($M_B = 10 M_S$). When we substitute this into the molality expression within the $\Delta T_b$ equation, the $M_S$ from the $K_b$ term perfectly cancels out the $M_S$ from the molality term.

Q: Can I approximate the solvent mass to 100 g to make calculations easier?

While approximating 99.75 g to 100 g might seem tempting and could give a close answer, JEE Advanced questions often have tight numerical tolerances. It's always safer to use the exact value, especially when the arithmetic simplifies nicely later on.

Question

In a solvent S, a compound B is partially dissociated into C and D as given below :
B $ightleftharpoons$ 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 $10R 	ext{ J mol}^{-1}$, respectively ($R$ is the gas constant in $	ext{J K}^{-1} 	ext{ mol}^{-1}$). A solution of B in S with an intial 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 _____.

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Boiling Point Elevation

Imagine you are in a chemistry lab, holding a beaker filled with a pure solvent, let's call it **S**. You measure its boiling point and find it to be exactly $400	ext{ K}$. Now, you dissolve a tiny amount of a mysterious compound **B** into it. The concentration is just $0.25\%$ by mass. You heat the solution again, and to your surprise, the boiling point has shot up to $408	ext{ K}$!

This is a classic case of **Elevation in Boiling Point**, a colligative property. The elevation, $\Delta T_b$, is simply the difference between the new boiling point and the original one:

$$\Delta T_b = 408	ext{ K} - 400	ext{ K} = 8	ext{ K}$$

### The Hidden Mechanics: Dissociation and van't Hoff Factor

But there's a twist. Compound **B** doesn't just sit there; it breaks apart! The reaction is given as:

$$B ightleftharpoons 2C + 2D$$

For every one molecule of **B** that dissociates, it produces $2$ molecules of **C** and $2$ molecules of **D**. That's a total of $n = 4$ particles. Because colligative properties depend on the *number* of particles, we must use the **van't Hoff factor ($i$)**. The formula linking $i$ to the degree of dissociation ($\alpha$) is:

$$i = 1 + (n - 1)\alpha$$

Substituting $n = 4$, we get:

$$i = 1 + 3\alpha$$

### Decoding the Concentration: Molality

Next, we need the molality ($m$) of the solution. The problem states we have a $0.25\% 	ext{ (w/w)}$ solution. This means in $100	ext{ g}$ of the total solution, there is $0.25	ext{ g}$ of solute **B**.

**Don't make a silly mistake here!** The mass of the solvent **S** is not $100	ext{ g}$, but rather $100 - 0.25 = 99.75	ext{ g}$. Let's plug this into the molality formula:

$$m = \frac{w_B 	imes 1000}{M_B 	imes W_S} = \frac{0.25 	imes 1000}{M_B 	imes 99.75} = \frac{1000}{399 M_B}$$

### The Thermodynamic Bridge: Ebullioscopic Constant ($K_b$)

We aren't given the ebullioscopic constant ($K_b$) directly. Instead, we must derive it using thermodynamics. The formula is:

$$K_b = \frac{R (T_b^\circ)^2 M_S}{1000 \Delta H_{vap}}$$

We know $T_b^\circ = 400	ext{ K}$ and $\Delta H_{vap} = 10R$. Substituting these values:

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

### The Master Equation: Bringing It All Together

Now, we assemble the master equation for boiling point elevation:

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

Substitute all the expressions we've carefully derived:

$$8 = (1 + 3\alpha) 	imes 16 M_S 	imes \frac{1000}{399 M_B}$$

Here is where the magic happens. The problem states that the molar mass of **B** is $10$ times that of **S**, so $M_B = 10 M_S$. Let's substitute this in:

$$8 = (1 + 3\alpha) 	imes 16 M_S 	imes \frac{1000}{399 	imes 10 M_S}$$

The unknown molar mass $M_S$ beautifully cancels out!

$$8 = (1 + 3\alpha) 	imes 16 	imes \frac{100}{399}$$

$$8 = (1 + 3\alpha) 	imes \frac{1600}{399}$$

### The Final Calculation

All that's left is some simple algebra to solve for $\alpha$:

$$1 + 3\alpha = \frac{8 	imes 399}{1600} = \frac{399}{200} = 1.995$$

$$3\alpha = 1.995 - 1 = 0.995$$

$$\alpha = \frac{0.995}{3} \approx 0.33166$$

To find the mole percent of **B** that has dissociated, we simply multiply $\alpha$ by $100$:

$$\%	ext{ Dissociation} = 0.33166 	imes 100 = 33.166\%$$

Rounding to two decimal places, we get our final, elegant answer: **$33.17\%$**.

Enter Numerical Value:

Visualized Solution

$Δ T_{b} = T_{b} - T_{b}^{\circ}$

$i = 1 + (n - 1) α$

$m = \frac{w _{B} \times 1000}{M _{B} \times W _{S}}$

$K_{b} = \frac{R ( T _{b}^{\circ} ) ^{2} M _{S}}{1000Δ H _{v a p}}$

$Δ T_{b} = i \cdot K_{b} \cdot m$

$M_{B} = 10 M_{S}$

$1 + 3 α = \frac{8 \times 399}{1600}$

$% Dissociation = α \times 100$

$The Way Forward$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Setup

The Hidden Mechanics

Decoding the Concentration

The Thermodynamic Bridge

The Master Equation

The Final Calculation

Similar Questions

A solute A dimerises in water. The boiling point of a 2 molar solution of A is $100.5 2^{\circ} C$ . The percentage association of A is ......... . (Round off to the nearest integer) [Use : $K_{b}$ for water $= 0.52 K kg mol^{- 1}$ , boiling point of water $= 10 0^{\circ} C$ ]

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1 molal aqueous solution of an electrolyte $A_{2} B_{3}$ is 60% ionised. The boiling point of the solution at 1 atm is ………… K (Rounded off to the nearest integer). [Given, $K_{b}$ for $(H_{2} O) = 0.52 K kg mol^{- 1}$ ]

The osmotic pressure of a dilute solution of an ionic compound $X Y$ in water is four times that of a solution of $0.01 M BaCl_{2}$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of $X Y$ (in $mol L^{- 1}$ ) in solution is

In a solvent 50% of an acid HA dimerises and the rest dissociates. The van't Hoff factor of the acid is ………… $\times 1 0^{- 2}$ . (Round off to the nearest integer)

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔTb​=Tb​−Tb∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=1+(n−1)α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=MB​×WS​wB​×1000​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kb​=1000ΔHvap​R(Tb∘​)2MS​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔTb​=i⋅Kb​⋅m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MB​=10MS​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+3α=16008×399​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } % Dissociation=α×100

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward