When 36.0 g of a solute having the empirical formula `CH_(2)O` is dissovled in 1.20 kg of water, the solution freezes at `-0.93^(@)C`. What is the moleculer formula of the solute ? (`K_(f) = 1.86^(@)C kg mol^(-1)`)

Options

A. `C_(2)H_(4)O`
B. `C_(2)H_(2)O_(2)`
C. `C_(2)H_(4)O_(3)`
D. `C_(2)H_(4)O_(2)`

Answer

To determine the molecular formula of the solute given its empirical formula (CH₂O) and the freezing point depression of the solution, we can follow these steps:

### Step 1: Calculate the depression in freezing point (ΔTf)
The depression in freezing point is given as:
\[ \Delta T_f = T_f^0 - T_f \]
Where:
- \( T_f^0 \) is the freezing point of pure water (0°C).
- \( T_f \) is the freezing point of the solution (-0.93°C).

Calculating ΔTf:
\[ \Delta T_f = 0 - (-0.93) = 0.93°C \]

### Step 2: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \( K_f \) is the cryoscopic constant (1.86°C kg/mol for water).
- \( m \) is the molality of the solution.

Rearranging the formula to find molality (m):
\[ m = \frac{\Delta T_f}{K_f} \]
Substituting the values:
\[ m = \frac{0.93}{1.86} \approx 0.5 \text{ mol/kg} \]

### Step 3: Calculate the number of moles of solute
The molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Given that the mass of the solvent (water) is 1.20 kg, we can rearrange to find the moles of solute:
\[ \text{moles of solute} = m \cdot \text{mass of solvent (kg)} = 0.5 \cdot 1.20 = 0.6 \text{ moles} \]

### Step 4: Calculate the molar mass of the solute
We know the mass of the solute is 36.0 g. To find the molar mass (M) of the solute, we use the formula:
\[ M = \frac{\text{mass of solute (g)}}{\text{moles of solute}} \]
Substituting the values:
\[ M = \frac{36.0 \text{ g}}{0.6 \text{ moles}} = 60 \text{ g/mol} \]

### Step 5: Determine the molecular formula
The empirical formula is CH₂O. To find the molecular formula, we need to determine how many times the empirical formula fits into the molar mass:
1. Calculate the molar mass of the empirical formula (CH₂O):
- C: 12 g/mol
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol
- Total = 12 + 2 + 16 = 30 g/mol

2. Now, divide the molar mass of the solute by the molar mass of the empirical formula:
\[ \text{n} = \frac{\text{Molar mass of solute}}{\text{Molar mass of empirical formula}} = \frac{60 \text{ g/mol}}{30 \text{ g/mol}} = 2 \]

3. Therefore, the molecular formula is:
\[ \text{Molecular formula} = (CH_2O)_2 = C_2H_4O_2 \]

### Final Answer
The molecular formula of the solute is **C₂H₄O₂**.

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