Pure solvent A has freezing point `16.5^(@)C`. On dissolving 0.4 g of B in 200 g of A, the solution freezing at `16.4^(@)C` and on dissolving 2.24 g of C in 100 g of A, the solution has freezing point of `16.0^(@)C`. If the molar mass of Bis `"74 g mol"^(-1),` what is the molar mass of C?

Explanation

To find the molar mass of substance C, we will follow these steps:

### Step 1: Calculate the change in freezing point (ΔTf) for solvent A when B is dissolved.
- The freezing point of pure solvent A is 16.5°C.
- The freezing point of the solution with B is 16.4°C.
- Therefore, ΔTf for B is:
\[ \Delta T_f = T_f(\text{pure}) - T_f(\text{solution}) = 16.5°C - 16.4°C = 0.1°C \]

### Step 2: Use the freezing point depression formula to find the molality (m) of B.
- The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \]
- Rearranging gives:
\[ m = \frac{\Delta T_f}{K_f} \]
- We know the mass of B (0.4 g) and the molar mass of B (74 g/mol). First, we need to find the number of moles of B:
\[ n_B = \frac{\text{mass}}{\text{molar mass}} = \frac{0.4 \text{ g}}{74 \text{ g/mol}} \approx 0.0054 \text{ mol} \]
- The mass of solvent A is 200 g or 0.2 kg. The molality (m) is:
\[ m = \frac{n_B}{\text{mass of solvent in kg}} = \frac{0.0054 \text{ mol}}{0.2 \text{ kg}} = 0.027 \text{ mol/kg} \]

### Step 3: Calculate the freezing point depression constant (Kf) using B.
- Now, we can find Kf:
\[ K_f = \frac{\Delta T_f}{m} = \frac{0.1°C}{0.027 \text{ mol/kg}} \approx 3.70 \text{ °C kg/mol} \]

### Step 4: Calculate the change in freezing point (ΔTf) for solvent A when C is dissolved.
- The freezing point of the solution with C is 16.0°C.
- Therefore, ΔTf for C is:
\[ \Delta T_f = 16.5°C - 16.0°C = 0.5°C \]

### Step 5: Use the freezing point depression formula to find the molality (m) of C.
- We can use the same formula:
\[ m_C = \frac{\Delta T_f}{K_f} = \frac{0.5°C}{3.70 \text{ °C kg/mol}} \approx 0.1351 \text{ mol/kg} \]

### Step 6: Calculate the number of moles of C.
- The mass of solvent A for C is 100 g or 0.1 kg. Thus, the number of moles of C is:
\[ n_C = m_C \times \text{mass of solvent in kg} = 0.1351 \text{ mol/kg} \times 0.1 \text{ kg} = 0.01351 \text{ mol} \]

### Step 7: Calculate the molar mass of C.
- The mass of C is given as 2.24 g. Therefore, the molar mass (M_C) of C is:
\[ M_C = \frac{\text{mass}}{\text{number of moles}} = \frac{2.24 \text{ g}}{0.01351 \text{ mol}} \approx 165.8 \text{ g/mol} \]

### Final Answer:
The molar mass of substance C is approximately **165.8 g/mol**.

---