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 \( C_2H_5OH \) (ethanol) in a water solution that freezes at \(-10^\circ C\), we can follow these steps:

### Step 1: Determine the change in freezing point (\( \Delta T_f \))
The freezing point of pure water is \( 0^\circ C \). The freezing point of the solution is given as \(-10^\circ C\).

\[ \Delta T_f = T_f(\text{solvent}) - T_f(\text{solution}) = 0^\circ C - (-10^\circ C) = 10^\circ C \]

### Step 2: Use the freezing point depression formula
The formula for freezing point depression is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \( \Delta T_f \) = change in freezing point
- \( i \) = van 't Hoff factor (which is 1 for non-electrolytes like ethanol)
- \( K_f \) = freezing point depression constant for water (given as \( 1.86 \, \text{°C kg/mol} \))
- \( m \) = molality of the solution

### Step 3: Substitute the known values into the formula
Since \( i = 1 \) for ethanol, we can simplify the equation:

\[ 10 = 1 \cdot 1.86 \cdot m \]

### Step 4: Solve for molality (\( m \))
Rearranging the equation to find \( m \):

\[ m = \frac{10}{1.86} \]

Calculating this gives:

\[ m \approx 5.38 \, \text{mol/kg} \]

### Final Answer
The molality of \( C_2H_5OH \) in the water solution that freezes at \(-10^\circ C\) is approximately \( 5.38 \, \text{mol/kg} \).

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