If 15gm of a solute in 100gm of water makes a solution that freezes at −1.0 o C then 30gm of the same solute in 100gm of water will make a solution that freezes at:

Explanation

To solve the problem, we need to determine the freezing point of a solution when 30 grams of solute is dissolved in 100 grams of water, given that 15 grams of the same solute in 100 grams of water freezes at -1.0 °C.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The freezing point depression (ΔTf) is directly proportional to the amount of solute in the solution. This means that if we double the amount of solute, the freezing point depression will also double.

2. **Given Information**:
- For 15 grams of solute, the freezing point (Tf1) is -1.0 °C.
- We need to find the freezing point (Tf2) when 30 grams of the same solute is used.

3. **Setting Up the Proportionality**:
We can express the relationship as:
\[ \frac{\Delta Tf1}{\Delta Tf2} = \frac{W1}{W2} \]
where:
- ΔTf1 is the freezing point depression for 15 grams of solute,
- ΔTf2 is the freezing point depression for 30 grams of solute,
- W1 = 15 grams,
- W2 = 30 grams.

4. **Calculating the Freezing Point Depression**:
Since the freezing point depression is the change in temperature from the pure solvent (0 °C for water), we can calculate:
- ΔTf1 = 0 °C - (-1.0 °C) = 1.0 °C.
- Therefore, we can set up the equation:
\[ \frac{1.0 °C}{\Delta Tf2} = \frac{15}{30} \]

5. **Solving for ΔTf2**:
Simplifying the right side:
\[ \frac{1.0 °C}{\Delta Tf2} = \frac{1}{2} \]
Cross-multiplying gives:
\[ \Delta Tf2 = 2.0 °C \]

6. **Finding the New Freezing Point**:
Since the freezing point depression for 30 grams of solute is 2.0 °C, we calculate the new freezing point:
\[ Tf2 = 0 °C - 2.0 °C = -2.0 °C \]

### Final Answer:
The solution that contains 30 grams of the same solute in 100 grams of water will freeze at -2.0 °C.