What is the molality of `C_2H_5 OH` in water solution which will freeze at `-10^@C`?

Options

A. 6.315m
B. 63.15m
C. 3.540m
D. 5.3m

Answer

To find the molality of ethanol (C₂H₅OH) in a water solution that freezes at -10°C, we can follow these steps:

### Step 1: Understand the Freezing Point Depression Formula
The freezing point depression (ΔTf) can be calculated using the formula:
\[ \Delta T_f = K_f \times m \]
where:
- \( \Delta T_f \) = change in freezing point
- \( K_f \) = freezing point depression constant of the solvent (water in this case)
- \( m \) = molality of the solution

### Step 2: Identify the Values
- The freezing point of pure water is 0°C.
- The freezing point of the solution is -10°C.
- Therefore, the change in freezing point (ΔTf) is:
\[ \Delta T_f = 0°C - (-10°C) = 10°C \]
- The freezing point depression constant \( K_f \) for water is given as 1.86 °C/m.

### Step 3: Substitute the Values into the Formula
Now we can substitute the values into the freezing point depression formula:
\[ 10°C = 1.86 \, \text{°C/m} \times m \]

### Step 4: Solve for Molality (m)
To find the molality (m), rearrange the equation:
\[ m = \frac{10°C}{1.86 \, \text{°C/m}} \]
Calculating this gives:
\[ m \approx 5.376 \, \text{molal} \]

### Step 5: Round to Appropriate Significant Figures
Since we typically round to two decimal places in chemistry, we can express the molality as:
\[ m \approx 5.38 \, \text{molal} \]

### Final Answer
The molality of ethanol in the water solution that freezes at -10°C is approximately **5.38 molal**.

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