A solution containing 6.8 g of non-ionic solute in 100g of water was found to freeze at `-0.93^(@)C`. If `K_(f)` for water is 1.86, the mol. Mass of solute is

Options

A. 13.6
B. 34
C. 68
D. 136

Explanation

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

### Step 1: Determine the change in freezing point (ΔTf)
The freezing point of pure water is 0°C. The freezing point of the solution is given as -0.93°C.

\[ \Delta T_f = T_f(\text{pure water}) - T_f(\text{solution}) = 0°C - (-0.93°C) = 0.93°C \]

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

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

Where:
- \( \Delta T_f \) is the change in freezing point (0.93°C),
- \( K_f \) is the cryoscopic constant (1.86°C kg/mol for water),
- \( m \) is the molality of the solution.

### Step 3: Rearranging the formula to find molality (m)
Rearranging the formula gives:

\[ m = \frac{\Delta T_f}{K_f} = \frac{0.93°C}{1.86°C \, \text{kg/mol}} = 0.5 \, \text{mol/kg} \]

### Step 4: Calculate the number of moles of solute
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We have 100 g of water, which is 0.1 kg.

\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \implies \text{moles of solute} = m \cdot \text{kg of solvent} = 0.5 \, \text{mol/kg} \cdot 0.1 \, \text{kg} = 0.05 \, \text{mol} \]

### Step 5: Calculate the molecular mass of the solute
The molecular mass (M) of the solute can be calculated using the formula:

\[ \text{Molecular mass} = \frac{\text{mass of solute (g)}}{\text{moles of solute}} = \frac{6.8 \, \text{g}}{0.05 \, \text{mol}} = 136 \, \text{g/mol} \]

### Final Answer
The molecular mass of the solute is **136 g/mol**.

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