20 g of a non-volatile solute is added to 500 g of solvent. Freezing point of solvent `= 5.48^(@)C` and solution `= 4.47^(@)C. K_(f)=1.93^(@)//m`. Molecular mas of the solute is

Options

A. `72.2`
B. `76.4`
C. `73.2`
D. `70.6`

Explanation

To find the molecular mass of the non-volatile solute, we can use the formula for depression in freezing point, which is a colligative property. Here's how to solve the problem step by step:

### Step 1: Determine the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:
\[ \Delta T_f = T_f^0 - T_f \]
where:
- \(T_f^0\) is the freezing point of the pure solvent (5.48°C)
- \(T_f\) is the freezing point of the solution (4.47°C)

Calculating ΔTf:
\[ \Delta T_f = 5.48°C - 4.47°C = 1.01°C \]

### Step 2: Use the depression in freezing point formula
The formula relating ΔTf to the molality (m) of the solution is:
\[ \Delta T_f = K_f \cdot m \]
where:
- \(K_f\) is the freezing point depression constant (1.93°C/m)
- \(m\) is the molality of the solution

Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} \]

Substituting the known values:
\[ m = \frac{1.01°C}{1.93°C/m} \approx 0.522 m \]

### Step 3: Calculate the number of moles of solute
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Given that the mass of the solvent is 500 g (or 0.5 kg), we can express the moles of solute as:
\[ \text{moles of solute} = m \cdot \text{mass of solvent in kg} = 0.522 \cdot 0.5 = 0.261 moles \]

### Step 4: Calculate the molecular mass of the solute
The molecular mass (M) of the solute can be calculated using the formula:
\[ \text{Molecular mass (M)} = \frac{\text{mass of solute}}{\text{moles of solute}} \]
Substituting the known values:
\[ M = \frac{20 g}{0.261 moles} \approx 76.6 g/mol \]

### Final Answer
The molecular mass of the solute is approximately 76.6 g/mol. The closest option provided in the question is 76.4 g/mol.

---