The freezing point of a diluted milk sample is found to be`-0.2^(@)C`, while it should have been `-0.5^(@)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

Explanation

To solve the problem, we need to determine how much water has been added to pure milk to create a diluted sample based on the freezing point depression.

### Step-by-Step Solution

1. **Understand the Freezing Point Depression**:
- The freezing point of pure milk is given as `-0.5°C`.
- The freezing point of the diluted milk sample is `-0.2°C`.
- The depression in freezing point (ΔTf) can be calculated as:
\[ \Delta T_f = T_f(\text{pure}) - T_f(\text{diluted}) = -0.5°C - (-0.2°C) = -0.3°C \]

2. **Use the Freezing Point Depression Formula**:
- The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Where:
- \(i\) = Van't Hoff factor (for non-electrolytes like milk, \(i = 1\))
- \(K_f\) = Freezing point depression constant (for water, \(K_f = 1.86 \, \text{°C kg/mol}\))
- \(m\) = Molality of the solution

3. **Set Up the Equations**:
- For the diluted milk sample:
\[ \Delta T_f = 1 \cdot K_f \cdot m_1 \quad \text{(Equation 1)} \]
- For pure milk:
\[ \Delta T_f = 1 \cdot K_f \cdot m_2 \quad \text{(Equation 2)} \]

4. **Calculate Molality**:
- The molality \(m\) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
- For the diluted milk sample, we can express it as:
\[ m_1 = \frac{W_m}{M_m \cdot W_1} \quad \text{(where \(W_m\) is the weight of milk, \(M_m\) is the molar mass of milk, and \(W_1\) is the weight of water)} \]
- For pure milk:
\[ m_2 = \frac{W_m}{M_m \cdot W_2} \quad \text{(where \(W_2\) is the weight of water in pure milk)} \]

5. **Set Up the Ratios**:
- From the two equations, we can set up a ratio:
\[ \frac{m_1}{m_2} = \frac{\Delta T_f \text{ (diluted)}}{\Delta T_f \text{ (pure)}} \]
- Substituting the values:
\[ \frac{m_1}{m_2} = \frac{0.3}{0.5} = \frac{3}{5} \]

6. **Determine the Amount of Water Added**:
- Let \(W_1\) be the weight of water added to the milk, and \(W_m\) be the weight of pure milk.
- From the ratio \( \frac{W_1}{W_m} = \frac{2}{3} \), we can conclude that for every 2 parts of water, there are 3 parts of milk.
- Therefore, if we have a total of 5 parts (2 parts water + 3 parts milk), we can say that the ratio of water to milk is 3:2.

### Conclusion
The amount of water added to pure milk is in the ratio of 3 cups of water to 2 cups of milk.