A 1 molal K4Fe(CN)6 solution has a degree of dissociation of 0.4. Its boiling point is equal to that

Q: Why do we equate the effective molality ($i \times m$) of both solutions?

The elevation in boiling point is given by $\Delta T_b = K_b \times i \times m$. Since both are aqueous solutions, $K_b$ is the same. If they boil at the same temperature, $\Delta T_b$ is the same. Therefore, the product $i \times m$ must be equal for both.

Q: How did we get $n=5$ for $K_4Fe(CN)_6$?

When $K_4Fe(CN)_6$ dissociates in water, it breaks down into four potassium ions ($4K^+$) and one complex ferrocyanide ion ($[Fe(CN)_6]^{4-}$). Adding these up gives a total of 5 ions per formula unit.

Q: Why is the van't Hoff factor ($i$) equal to 1 for Solute A?

The problem states that Solute A is a 'non-electrolytic solute'. Non-electrolytes do not dissociate into ions or associate into larger molecules in solution. Therefore, one molecule of solute yields exactly one particle in solution, making $i = 1$.

Q: What exactly does '18.1 weight per cent' mean?

It means that in every 100 grams of the total solution, there are 18.1 grams of the solute. The rest of the mass is the solvent.

Q: Why did we subtract 18.1 from 100 to find the mass of the solvent?

Molality is defined as the moles of solute divided by the mass of the *solvent* (not the total solution) in kilograms. If we have 100 g of solution and 18.1 g is solute, the remaining mass ($100 - 18.1 = 81.9$ g) must be the water (solvent).

Question

A 1 molal $K_4Fe(CN)_6$ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight per cent of a non-electrolytic solute A. The molar mass of A is ....... u. (Round off to the nearest integer).
[Density of water = $1.0 	ext{ g cm}^{-3}$]

Sigma Percentile · Accepted Answer

### The Tale of Two Beakers

Imagine you are in a chemistry lab, and you have two beakers bubbling away on hot plates. In the first beaker, you have a $1$ molal solution of a complex salt, $K_4Fe(CN)_6$. In the second beaker, you have an unknown non-electrolytic solute, let's call it Solute A, dissolved in water to make an $18.1\%$ by weight solution.

You glance at the thermometers in both beakers and notice something fascinating: they are both boiling at the exact same temperature! This isn't just a coincidence; it's a beautiful demonstration of colligative properties in action. Let's dive into the math and uncover the molar mass of our mystery Solute A.

### The Master Equation: Equating Effective Molality

We know that the elevation in boiling point, $\Delta T_b$, is a colligative property. It depends only on the number of solute particles, not their identity. The formula is:

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

Since both solutions are aqueous, they share the same boiling point elevation constant, $K_b$. Because they boil at the same temperature, their $\Delta T_b$ values are identical. This leads us to a powerful conclusion: their **effective molalities** must be equal.

$$i_1 \cdot m_1 = i_2 \cdot m_2$$

### Analyzing the Complex Salt

Let's look at our first beaker containing $K_4Fe(CN)_6$. This salt dissociates in water:

$$K_4Fe(CN)_6 ightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$$

One molecule yields $5$ ions, so $n = 5$. However, it doesn't dissociate completely; its degree of dissociation, $\alpha$, is $0.4$. We can calculate its van't Hoff factor, $i_1$:

$$i_1 = 1 + (n - 1)\alpha$$
$$i_1 = 1 + (5 - 1) 	imes 0.4 = 1 + 1.6 = 2.6$$

Since its stated molality $m_1$ is $1$ m, its effective molality is:

$$i_1 \cdot m_1 = 2.6 	imes 1 = 2.6 	ext{ m}$$

### Unveiling the Mystery Solute

Now, let's turn our attention to the second beaker. Solute A is a non-electrolyte, which means it doesn't break apart into ions. Therefore, its van't Hoff factor, $i_2$, is simply $1$.

To match the effective molality of the first beaker, the actual molality of Solute A must also be $2.6$ m. We are given that it's an $18.1\%$ by weight solution. This means in $100$ g of the solution, there are $18.1$ g of Solute A and $100 - 18.1 = 81.9$ g of water (the solvent).

Let's set up the molality equation for Solute A:

$$m_2 = \frac{	ext{Moles of Solute A}}{	ext{Mass of solvent in kg}}$$
$$2.6 = \frac{18.1 / M_A}{81.9 / 1000}$$

### The Final Calculation

All that's left is some careful algebra to isolate $M_A$, the molar mass of Solute A:

$$2.6 = \frac{18.1 	imes 1000}{M_A 	imes 81.9}$$
$$M_A = \frac{18100}{2.6 	imes 81.9}$$
$$M_A = \frac{18100}{212.94} \approx 85 	ext{ g/mol}$$

And there we have it! By understanding how particles behave in a solution and equating their colligative effects, we successfully deduced that the molar mass of our mystery Solute A is **$85$ g/mol**.

Enter Numerical Value:

Visualized Solution

$T_{b 1} = T_{b 2}$

$Δ T_{b} = i \cdot K_{b} \cdot m$

$i_{1} \cdot m_{1} = 2.6$

$m_{2} = \frac{W _{A} / M _{A}}{W _{so l v e n t} /1000}$

$2.6 = \frac{18.1 \times 1000}{M _{A} \times 81.9}$

$M_{A} = 85 g/mol$

$What if A was an electrolyte?$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Tale of Two Beakers

The Master Equation

Analyzing the Complex Salt

Unveiling the Mystery Solute

The Final Calculation

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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; } Tb1​=Tb2​

.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; } i1​⋅m1​=2.6

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​=Wsolvent​/1000WA​/MA​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2.6=MA​×81.918.1×1000​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MA​=85 g/mol

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if A was an electrolyte?