If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous sol

Q: Why do we multiply by 3? Isn't the moles given as 0.01?

We multiply by 3 because sodium sulphate is an electrolyte that breaks apart in water. One molecule of $\text{Na}_2\text{SO}_4$ splits into two $\text{Na}^+$ ions and one $\text{SO}_4^{2-}$ ion. Colligative properties depend on the total number of particles, so we must account for this using the van't Hoff factor ($i=3$).

Q: What if the question didn't say 'completely dissociated'?

If it wasn't completely dissociated, it would be a weak electrolyte. You would then need the degree of dissociation ($\alpha$) to calculate the van't Hoff factor using the formula $i = 1 + (n-1)\alpha$, where $n$ is the maximum number of ions it can form.

Q: Why do we use molality instead of molarity here?

Molality is used in colligative properties involving temperature changes (like boiling point elevation and freezing point depression) because molality is mass-based and does not change with temperature. Molarity is volume-based, and volume expands or contracts with temperature changes.

Q: Is the unit of $\Delta T_f$ in Kelvin or Celsius?

A change in temperature ($\Delta T$) is the same magnitude whether it is measured in Kelvin or Celsius. So, a drop of $0.0558\text{ K}$ is exactly the same as a drop of $0.0558^\circ\text{C}$.

Q: Can I use this formula for any solute?

Yes, but you must be careful with the van't Hoff factor ($i$). For non-electrolytes like glucose or urea that don't dissociate, $i = 1$. For electrolytes, $i$ depends on how many ions they form.

Question

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water ($\Delta T_f$), when $0.01$ mole of sodium sulphate is dissolved in $1	ext{ kg}$ of water, is ($K_f = 1.86	ext{ K kg mol}^{-1}$)

Options:

(A) $0.0372	ext{ K}$

(B) $0.0558	ext{ K}$

(C) $0.0744	ext{ K}$

(D) $0.0186	ext{ K}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing in a laboratory with a beaker containing exactly $1	ext{ kg}$ of pure water. You are handed $0.01	ext{ moles}$ of sodium sulphate ($	ext{Na}_2	ext{SO}_4$) and asked to predict how much the freezing point of the water will drop once you dissolve the salt in it.

This is a classic problem of colligative properties, specifically the **depression in freezing point**. Colligative properties depend entirely on the *number* of solute particles in the solution, not on their chemical identity.

### The Master Equation
To find the change in freezing point, we rely on the fundamental equation:

$$\Delta T_f = i \cdot K_f \cdot m$$

Here, $\Delta T_f$ is the depression in freezing point, $i$ is the van't Hoff factor, $K_f$ is the molal depression constant (or cryoscopic constant) of the solvent, and $m$ is the molality of the solution.

### Decoding the van't Hoff Factor ($i$)
The most critical part of this problem is recognizing that sodium sulphate is a **strong electrolyte**. When it dissolves in water, it doesn't stay as intact $	ext{Na}_2	ext{SO}_4$ molecules. Instead, it completely dissociates into its constituent ions:

$$	ext{Na}_2	ext{SO}_4 ightarrow 2	ext{Na}^+ + 	ext{SO}_4^{2-}$$

For every one formula unit of sodium sulphate that dissolves, we get two sodium ions and one sulphate ion, making a total of **three particles**. Because the problem states it is completely dissociated, our van't Hoff factor $i$ is exactly $3$.

### Calculating Molality ($m$)
Next, we need the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

$$m = \frac{	ext{Moles of solute}}{	ext{Mass of solvent (in kg)}}$$

Substituting our given values:

$$m = \frac{0.01	ext{ mol}}{1	ext{ kg}} = 0.01	ext{ mol kg}^{-1}$$

### Final Calculation
Now, we have all the pieces of the puzzle. We know $i = 3$, $K_f = 1.86	ext{ K kg mol}^{-1}$, and $m = 0.01	ext{ mol kg}^{-1}$. Let's plug them into our master equation:

$$\Delta T_f = 3 	imes 1.86 	imes 0.01$$

First, multiply $3$ by $1.86$:

$$3 	imes 1.86 = 5.58$$

Finally, multiply by $0.01$:

$$\Delta T_f = 5.58 	imes 0.01 = \mathbf{0.0558	ext{ K}}$$

The freezing point of the water will drop by $0.0558	ext{ K}$. This elegant calculation shows how a tiny amount of salt can measurably alter the physical properties of a solvent!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Dissociation and van’t Hoff Factor$

$Calculating Molality$

$Freezing Point Depression Formula$

$Substituting Values$

$Executing the Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Colligative Properties

Analyzing the Setup Imagine you are standing in a laboratory with a beaker containing exactly $1 kg$ of pure water

The Master Equation

Decoding the van't Hoff Factor ( $i$ ) The most critical part of this problem is recognizing that sodium sulphate is a strong electrolyte

Calculating Molality ( $m$ ) Next, we need the molality of the solution

Final Calculation Now, we have all the pieces of the puzzle

Similar Questions

Molal depression constant for a solvent is $4.0 K kg mol^{- 1}$ . The depression in the freezing point of the solvent for $0.03 mol kg^{- 1}$ solution of $K_{2} SO_{4}$ is (Assume complete dissociation of the electrolyte)

How much amount of NaCl should be added to 600 g of water ( $ρ = 1.00 g/mL$ ) to decrease the freezing point of water to $- 0. 2^{\circ} C$ ? ...... (The freezing point depression constant for water $= 2 K kg mol^{- 1}$ )

For an aqueous solution, freezing point is $- 0.18 6^{\circ} C$ . Elevation of the boiling point of the same solution is ( $K_{f} = 1.8 6^{\circ} mol^{- 1} kg$ and $K_{b} = 0.51 2^{\circ} mol^{- 1} kg$ )

$K_{f}$ for water is $1.86 K kg mol^{- 1}$ . If your automobile radiator holds $1.0 kg$ of water, then how many grams of ethylene glycol ( $C_{2} H_{6} O_{2}$ ) must you add to get the freezing point of the solution lowered to $- 2. 8^{\circ} C$ ?

A solution contain $62 g$ of ethylene glycol in $250 g$ of water is cooled upto $- 1 0^{\circ} C$ . If $K_{f}$ for water is $1.86 K kg mol^{- 1}$ , then amount of water (in g) separated as ice is

If the freezing point of a $0.01$ molal aqueous solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong electrolyte) is $- 0.055 8^{\circ} C$ , the number of chloride(s) in the coordination sphere of the complex is [ $K_{f}$ of water = $1.86 K kg mol^{- 1}$ ]

Which one of the following 0.06 M aqueous solutions has lowest freezing point? [2021, 22 July Shift-II]

83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ………K. (Nearest integer) [Use, molal freezing point depression constant of water $= 1.86 K kg mol^{- 1}$ , Freezing point of water $= 273 K$ and Atomic masses : $C = 12.0 u, O = 16.0 u, H = 1.0 u$ ]

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to $4 kg$ of water to prevent it from freezing at $- 6^{\circ} C$ will be ( $K_{f}$ for water $= 1.86 K kg mol^{- 1}$ and molar mass of ethylene glycol $= 62 g mol^{- 1}$ )

Elevation in the boiling point for $1$ molal solution of glucose is $2 K$ . The depression in the freezing point for $2$ molal solution of glucose in the same solvent is $2 K$ . The relation between $K_{b}$ and $K_{f}$ is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dissociation and van’t Hoff Factor

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Freezing Point Depression Formula

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Executing the Calculation

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

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