The freezing point of a solution prepared from `1.25 g` of non-electrolyte and `20 g` of water is `271.9 K`. If the molar depression constant is `1.86 K mol^(-1)`, then molar mass of the solute will be

Options

A. 105.7
B. 106.7
C. 115.3
D. 93.9

Explanation

To find the molar mass of the solute, we can follow these steps:

### Step 1: Calculate the depression in freezing point (ΔTf)
The freezing point of pure water (Tf0) is 273 K. The freezing point of the solution (TfS) is given as 271.9 K.

\[ \Delta Tf = Tf0 - TfS = 273 K - 271.9 K = 1.1 K \]

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

\[ \Delta Tf = Kf \cdot m \]

where:
- \( Kf \) is the molal depression constant (1.86 K kg/mol),
- \( m \) is the molality of the solution.

### Step 3: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:

\[ m = \frac{n}{W_{a}} \]

where:
- \( n \) is the number of moles of solute,
- \( W_{a} \) is the mass of the solvent in kg.

Given that the mass of water (solvent) is 20 g, we convert it to kg:

\[ W_{a} = 20 \, \text{g} = 0.020 \, \text{kg} \]

### Step 4: Substitute values into the freezing point depression formula
We can rearrange the freezing point depression formula to find the number of moles of solute (n):

\[ \Delta Tf = Kf \cdot \frac{n}{W_{a}} \]

Rearranging gives:

\[ n = \frac{\Delta Tf \cdot W_{a}}{Kf} \]

Substituting the known values:

\[ n = \frac{1.1 \, K \cdot 0.020 \, kg}{1.86 \, K \cdot kg/mol} \]

Calculating this gives:

\[ n = \frac{0.022}{1.86} \approx 0.01183 \, \text{mol} \]

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

\[ M = \frac{W_{b}}{n} \]

where:
- \( W_{b} \) is the mass of the solute (1.25 g),
- \( n \) is the number of moles of solute.

Substituting the values:

\[ M = \frac{1.25 \, g}{0.01183 \, mol} \approx 105.68 \, g/mol \]

### Final Answer
The molar mass of the solute is approximately **105.68 g/mol**.

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