The freezing point of a diluted milk sample is found to be -0.2°C, while it should have been -0.5°C

Q: Why does adding water increase the freezing point from $-0.5^{\circ}\text{C}$ to $-0.2^{\circ}\text{C}$?

Pure water freezes at $0^{\circ}\text{C}$. When solutes are present (like in pure milk), the freezing point drops to $-0.5^{\circ}\text{C}$. Adding more water dilutes the solute concentration, making the solution behave more like pure water, so the freezing point moves closer to $0^{\circ}\text{C}$ (i.e., it increases to $-0.2^{\circ}\text{C}$).

Q: Why did we assume the van't Hoff factor $i$ is 1?

Milk is a complex mixture, but the primary solutes causing the freezing point depression (like lactose) are non-electrolytes. Even if there are some salts, their effective $i$ remains constant in both pure and diluted milk, so it cancels out when we take the ratio.

Q: How did we get the ratio of 3 cups to 2 cups from $2/5$?

The ratio $w_2/w_1 = 2/5$ means that for every 2 parts of water originally in the pure milk ($w_2$), there are now 5 parts of water in the diluted milk ($w_1$). The amount of water added is $w_1 - w_2 = 5 - 2 = 3$ parts. So, 3 parts were added to the original 2 parts.

Q: Do we need to know the exact mass of the milk solids to solve this?

Not at all! That's the beauty of taking a ratio. The mass of the milk solids ($w_{\text{solute}}$) and its molar mass ($M_{\text{solute}}$) remain constant during dilution, so they completely cancel out when you divide the two equations.

Q: Can we use volume instead of mass for the cups of water?

Yes, because the density of water is approximately $1 \text{ g/mL}$, the mass ratio directly translates to a volume ratio. So, a mass ratio of 3:2 is equivalent to a volume ratio of 3 cups to 2 cups.

Question

The freezing point of a diluted milk sample is found to be $-0.2^{\circ}	ext{C}$, while it should have been $-0.5^{\circ}	ext{C}$ for pure milk. How much water has been added to pure milk to make the diluted sample?

Options:

(A) 2 cups of water to 3 cups of pure milk

(B) 1 cup of water to 3 cups of pure milk

(C) 3 cups of water to 2 cups of pure milk

(D) 1 cup of water to 2 cups of pure milk

Sigma Percentile · Accepted Answer

### Unmasking the Milkman: A Tale of Freezing Points

Imagine you are a food inspector tasked with a very important mission: finding out if the local milkman has been secretly adding water to his milk. You don't have a complex chemical laboratory, but you do have a thermometer. You measure the freezing point of a sample and find it to be $-0.2^{\circ}	ext{C}$. However, you know that pure milk should freeze at $-0.5^{\circ}	ext{C}$. How can you use this simple temperature difference to calculate exactly how much water was added?

This is where the magic of **colligative properties** comes into play.

### The Science of Freezing Point Depression

Milk is essentially an aqueous solution. It contains water as the solvent and various milk solids (like lactose, proteins, and fats) acting as the solute. When a non-volatile solute is dissolved in a solvent, it lowers the freezing point of the solvent. This phenomenon is known as **depression in freezing point**.

The mathematical relationship is given by the formula:

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

Here, $\Delta T_f$ is the depression in freezing point ($T_f^{\circ} - T_f$), $K_f$ is the molal depression constant of the solvent (water), $m$ is the molality of the solution, and $i$ is the van't Hoff factor.

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent. We can expand it as:

$$m = \frac{w_{	ext{solute}} 	imes 1000}{M_{	ext{solute}} 	imes w_{	ext{solvent}}}$$

### Setting up the Math

Let's apply this to our two scenarios: the pure milk and the diluted milk.

For the **pure milk**, the freezing point is $-0.5^{\circ}	ext{C}$, so the depression $\Delta T_{f2} = 0 - (-0.5) = 0.5^{\circ}	ext{C}$. Let the mass of water in pure milk be $w_2$.

$$0.5 = K_f 	imes \frac{w_{	ext{milk}} 	imes 1000}{M_{	ext{milk}} 	imes w_2} 	imes 1$$

For the **diluted milk**, the freezing point is $-0.2^{\circ}	ext{C}$, so the depression $\Delta T_{f1} = 0 - (-0.2) = 0.2^{\circ}	ext{C}$. Let the mass of water in the diluted milk be $w_1$.

$$0.2 = K_f 	imes \frac{w_{	ext{milk}} 	imes 1000}{M_{	ext{milk}} 	imes w_1} 	imes 1$$

Notice a crucial detail here: when the milkman adds water, the amount of milk solids ($w_{	ext{milk}}$) does **not** change. Only the mass of the solvent (water) increases.

### The Elegant Cancellation

We have two equations with several unknown variables like $K_f$, $w_{	ext{milk}}$, and $M_{	ext{milk}}$. But we don't need to know them! By dividing the equation for the diluted milk by the equation for the pure milk, all these constant terms beautifully cancel out.

$$\frac{0.2}{0.5} = \frac{K_f 	imes \frac{w_{	ext{milk}} 	imes 1000}{M_{	ext{milk}} 	imes w_1}}{K_f 	imes \frac{w_{	ext{milk}} 	imes 1000}{M_{	ext{milk}} 	imes w_2}}$$

This simplifies elegantly to:

$$\frac{2}{5} = \frac{w_2}{w_1}$$

### Interpreting the Final Ratio

What does this ratio tell us? It says that the mass of water in the pure milk ($w_2$) is to the mass of water in the diluted milk ($w_1$) as $2$ is to $5$.

If the pure milk originally contained $2$ parts of water, the diluted milk now contains $5$ parts of water. This means the milkman added $5 - 2 = 3$ parts of water to the original $2$ parts of pure milk.

Therefore, the milkman added **3 cups of water to every 2 cups of pure milk**. Science catches the culprit once again!

Animated Solution for Chemistry - Solutions: The freezing point of a diluted milk sample is found to be $- 0. 2^{\circ} C$ , while it should have been $- 0. 5^{\circ} C$ for pure milk. How much water has been added to pure milk to make the diluted sample?

Select Answer:

Visualized Solution

\text{The Adulteration Problem}

\text{Depression in Freezing Point}

\text{Setting up the Equations}

\text{Taking the Ratio}

\text{Interpreting the Ratio}

\text{Conclusion}

The Sigma Insight: Colligative Properties

Unmasking the Milkman

The Science of Freezing Point Depression

Setting up the Math

The Elegant Cancellation

Interpreting the Final Ratio

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