A solution containing 18 g of a non - electrolytic solute in 200g of `H_(2)O` freezes at 272.07 K . Find the molecular mass of the solute . (`K_(f) = 1.86 Km^(-1)`).

Explanation

To find the molecular mass of the solute in the given solution, we can follow these steps:

### Step 1: Identify the known values
- Mass of solute (non-electrolytic) = 18 g
- Mass of solvent (water) = 200 g = 0.200 kg (convert grams to kilograms)
- Freezing point of the solution = 272.07 K
- Freezing point of pure water = 0 °C = 273.15 K
- Kf (freezing point depression constant for water) = 1.86 K kg/mol

### Step 2: Calculate the depression in freezing point (ΔTf)
\[ \Delta T_f = T_{initial} - T_{final} \]
\[ \Delta T_f = 273.15 \, \text{K} - 272.07 \, \text{K} = 1.08 \, \text{K} \]

### Step 3: 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-electrolytic solute, \(i = 1\))
- \(K_f\) = freezing point depression constant
- \(m\) = molality of the solution

### Step 4: Calculate molality (m)
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
To find the moles of solute, we use:
\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{18 \, \text{g}}{M} \]
Thus, molality can be expressed as:
\[ m = \frac{18/M}{0.200} = \frac{90}{M} \, \text{mol/kg} \]

### Step 5: Substitute values into the freezing point depression formula
Substituting \(i\), \(K_f\), and \(m\) into the formula:
\[ 1.08 = 1 \cdot 1.86 \cdot \frac{90}{M} \]

### Step 6: Solve for M (molar mass of solute)
Rearranging the equation gives:
\[ M = \frac{1.86 \cdot 90}{1.08} \]
Calculating this:
\[ M = \frac{167.4}{1.08} \approx 155 \, \text{g/mol} \]

### Final Answer
The molecular mass of the solute is approximately **155 g/mol**.

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