How much amount of NaCl should be added to 600 g of water (rho = 1.00 g/mL) to decrease the freezing

Q: Why do we use $i=2$ for NaCl?

NaCl is a strong electrolyte. When dissolved in water, it completely dissociates into two separate ions: one $\text{Na}^+$ and one $\text{Cl}^-$. The van't Hoff factor ($i$) represents the number of particles formed in solution, so $i=2$.

Q: Why is $\Delta T_f$ taken as positive $0.2$ and not $-0.2$?

$\Delta T_f$ represents the *depression* or the *change* in the freezing point. It is defined as $T_f^\circ - T_f$ (Freezing point of pure solvent minus freezing point of solution). So, $0^\circ\text{C} - (-0.2^\circ\text{C}) = +0.2^\circ\text{C}$ (or $0.2 \text{ K}$).

Q: Does the density of water ($\rho = 1.00 \text{ g/mL}$) matter here?

In this specific problem, the mass of water is already given directly as $600 \text{ g}$. The density would only be needed if the volume of water was given instead of its mass. Here, it's just extra information.

Q: What if the solute was glucose instead of NaCl?

Glucose is a non-electrolyte, meaning it does not dissociate into ions in water. For glucose, the van't Hoff factor $i$ would be $1$. You would need twice as many moles of glucose to achieve the same freezing point depression as NaCl.

Q: Why do we multiply by 1000 in the molality formula?

Molality is defined as moles of solute per *kilogram* of solvent. Since the mass of the solvent ($w_A$) is given in grams ($600 \text{ g}$), we multiply by $1000$ to convert those grams into kilograms within the formula.

Question

How much amount of NaCl should be added to 600 g of water ($ho = 1.00 	ext{ g/mL}$) to decrease the freezing point of water to $-0.2^\circ	ext{C}$? ...... (The freezing point depression constant for water $= 2 	ext{ K kg mol}^{-1}$)

Sigma Percentile · Accepted Answer

### The Magic of Freezing Point Depression

Imagine you have a beaker filled with exactly $600 	ext{ g}$ of pure water. We all know that pure water freezes at exactly $0^\circ	ext{C}$. But what if we want to keep it liquid just a little bit longer? What if we want to push that freezing point down to $-0.2^\circ	ext{C}$?

This isn't just a theoretical exercise; it's the exact reason why salt trucks roam the streets during winter storms. By adding a solute to a pure solvent, we disrupt the solvent's ability to form a solid crystal lattice, effectively lowering the temperature required for it to freeze. This phenomenon is known as **Depression in Freezing Point**, and it is a classic colligative property.

### Unpacking the Master Equation

To figure out exactly how much salt (NaCl) we need, we rely on the master equation for freezing point depression:

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

Let's break this down. $\Delta T_f$ is the change in the freezing point. Since we are going from $0^\circ	ext{C}$ to $-0.2^\circ	ext{C}$, our $\Delta T_f$ is a positive $0.2 	ext{ K}$.

The term $K_f$ is the cryoscopic constant, a unique property of the solvent. For water, the problem generously provides $K_f = 2 	ext{ K kg mol}^{-1}$.

The term $m$ stands for **molality**, which is the number of moles of solute per kilogram of solvent. We can expand molality into a more useful form:

$$ m = \frac{w_B 	imes 1000}{M_B 	imes w_A} $$

Here, $w_B$ is the mass of our solute (NaCl) that we want to find, $M_B$ is its molar mass, and $w_A$ is the mass of our solvent (water) in grams.

### The Power of the van't Hoff Factor

Now, let's talk about the most crucial part of this equation: the van't Hoff factor, **$i$**.

Sodium chloride is a strong electrolyte. When you drop it into water, it doesn't just sit there as intact NaCl molecules. It completely dissociates into two distinct ions: one sodium ion ($	ext{Na}^+$) and one chloride ion ($	ext{Cl}^-$).

Because colligative properties depend strictly on the *number* of particles in solution, not their identity, one mole of NaCl effectively acts like two moles of particles. Therefore, for NaCl, **$i = 2$**. This is a massive advantage; it means salt is twice as effective at melting ice as a non-electrolyte like sugar!

### Crunching the Numbers

We have all our pieces. Let's assemble the puzzle. We know the molar mass of NaCl ($M_B$) is $23 + 35.5 = 58.5 	ext{ g mol}^{-1}$. Let's substitute everything into our expanded master equation:

$$ \Delta T_f = \frac{i \cdot K_f \cdot w_B \cdot 1000}{M_B \cdot w_A} $$

$$ 0.2 = \frac{2 \cdot 2 \cdot w_B \cdot 1000}{58.5 \cdot 600} $$

Now, it's just a matter of algebra. We need to isolate $w_B$:

$$ w_B = \frac{0.2 \cdot 58.5 \cdot 600}{2 \cdot 2 \cdot 1000} $$

Let's simplify the numerator and denominator:

$$ w_B = \frac{7020}{4000} $$

$$ w_B = 1.755 	ext{ g} $$

### The Final Result

Rounding our answer to two decimal places, we find that we need exactly **$1.76 	ext{ g}$** of NaCl.

Think about that for a second. Just $1.76$ grams of salt—barely a pinch—is enough to alter the physical state of over half a kilogram of water. That is the sheer power of chemistry and colligative properties at work!

Animated Solution for Chemistry - Solutions: 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 $= 2 K kg mol^{- 1}$ )

Enter Numerical Value:

Visualized Solution

\text{Understanding the Setup}

\text{Formula for Depression in Freezing Point}

\text{Identifying the Parameters}

\text{Substituting the Values}

\text{Calculating the Mass of NaCl}

\text{Conceptual Takeaway}

The Sigma Insight: Colligative Properties

The Magic of Freezing Point Depression

Unpacking the Master Equation

The Power of the van't Hoff Factor

Crunching the Numbers

The Final Result

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